If $a+b=1$ find the greatest value for $a^2b^3$
I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited yesterday
Michael Rozenberg
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asked yesterday
user587054
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I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited yesterday
Michael Rozenberg
97.2k1589188
asked yesterday
user587054
47211
4
I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday
add a comment |
I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited yesterday
Michael Rozenberg
97.2k1589188
asked yesterday
user587054
47211
I have been trying for some time on this question but i am new to inequalities so I am unable to solve it. I tried am gm but failed. Any help would be apriciated
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited yesterday
Michael Rozenberg
97.2k1589188
asked yesterday
user587054
47211
edited yesterday
Michael Rozenberg
97.2k1589188
asked yesterday
user587054
47211
edited yesterday
Michael Rozenberg
97.2k1589188
edited yesterday
Michael Rozenberg
97.2k1589188
edited yesterday
Michael Rozenberg
97.2k1589188
97.2k1589188
asked yesterday
user587054
47211
asked yesterday
user587054
47211
asked yesterday
user587054
47211
47211
4
I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday
add a comment |
4
I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday
4
4
I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday
I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday
add a comment |
5 Answers
5
active
oldest
votes
Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.
answered yesterday
Lord Shark the Unknown
102k958132
add a comment |
Just using algebra.
Using $b=1-a$, you are looking for the maximum of function
$$f(a)=a^2(1-a)^3$$ Compute the derivatives
$$f'(a)=-a(1-a)^2 (5 a-2)$$
$$f''(a)=2(1-a)(10 a^2-8 a+1)$$
The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.
answered yesterday
Claude Leibovici
119k1157132
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
add a comment |
Does not exist. Try $brightarrow+infty$.
For non-negatives $a$ and $b$ by AM-GM we obtain:
$$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.
answered yesterday
Michael Rozenberg
97.2k1589188
add a comment |
$$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
$$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
$$f'(x) = 0 implies x in {0, 1, frac 35}$$
$$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
$$f''(0)=0, f''(1)>0, f''(frac 35)<0$$
Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$
answered yesterday
Rhys Hughes
5,0441427
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
add a comment |
Let Constraint function be
$$C(a,b)= a+b-1=0 $$
and the Object function that needs maximization be
$$G(a,b)=a^2b^3 $$
we have with partial differentiation of combined Lagrangian:
$$ C(a,b)- lambda G(a,b) $$
condition to evaluate the multiplier $lambda$
$$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
$$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$
That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.
After plugging in these values Object function maximum value is:
$$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$
answered yesterday
Narasimham
20.6k52158
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.
answered yesterday
Lord Shark the Unknown
102k958132
add a comment |
Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.
answered yesterday
Lord Shark the Unknown
102k958132
add a comment |
Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.
answered yesterday
Lord Shark the Unknown
102k958132
Assuming that $a$ and $b$ are nonnegative, apply AM/GM to $a/2$, $a/2$, $b/3$, $b/3$ and $b/3$ which sum to $1$. One gets
$$frac15gesqrt[5]{frac a2frac a2frac b3frac b3frac b3}$$
etc.
answered yesterday
Lord Shark the Unknown
102k958132
answered yesterday
Lord Shark the Unknown
102k958132
answered yesterday
Lord Shark the Unknown
102k958132
answered yesterday
Lord Shark the Unknown
102k958132
102k958132
add a comment |
add a comment |
Just using algebra.
Using $b=1-a$, you are looking for the maximum of function
$$f(a)=a^2(1-a)^3$$ Compute the derivatives
$$f'(a)=-a(1-a)^2 (5 a-2)$$
$$f''(a)=2(1-a)(10 a^2-8 a+1)$$
The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.
answered yesterday
Claude Leibovici
119k1157132
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
add a comment |
Just using algebra.
Using $b=1-a$, you are looking for the maximum of function
$$f(a)=a^2(1-a)^3$$ Compute the derivatives
$$f'(a)=-a(1-a)^2 (5 a-2)$$
$$f''(a)=2(1-a)(10 a^2-8 a+1)$$
The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.
answered yesterday
Claude Leibovici
119k1157132
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
add a comment |
Just using algebra.
Using $b=1-a$, you are looking for the maximum of function
$$f(a)=a^2(1-a)^3$$ Compute the derivatives
$$f'(a)=-a(1-a)^2 (5 a-2)$$
$$f''(a)=2(1-a)(10 a^2-8 a+1)$$
The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.
answered yesterday
Claude Leibovici
119k1157132
Just using algebra.
Using $b=1-a$, you are looking for the maximum of function
$$f(a)=a^2(1-a)^3$$ Compute the derivatives
$$f'(a)=-a(1-a)^2 (5 a-2)$$
$$f''(a)=2(1-a)(10 a^2-8 a+1)$$
The first derivative cancels at $a=0$, $a=1$ and $a=frac 25$. For the first two, $f(a)=0$. For the last one $fleft(frac{2}{5}right)=frac{108}{3125}$ and $f''left(frac{2}{5}right)=-frac{18}{25} <0$ confirms that this is a maximum value.
answered yesterday
Claude Leibovici
119k1157132
answered yesterday
Claude Leibovici
119k1157132
answered yesterday
Claude Leibovici
119k1157132
answered yesterday
Claude Leibovici
119k1157132
119k1157132
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
add a comment |
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
Although it's not the most elegant, this is by far the most straightforward answer!
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
You might want to add in the intermediate steps $f'(a) = -3a^2(1-a)^2$ and $f''(a) = -(1-a)^2(5a-2) + -a(1-a)(-2)(5a-2) - a(1-a)^2(5)$ to make reading easier.
– Toby Mak
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
Is $f''(x)$ required at all?
– Shubham Johri
yesterday
2
2
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
@ShubhamJohri; Just to confirm that this is a maximum.
– Claude Leibovici
yesterday
add a comment |
Does not exist. Try $brightarrow+infty$.
For non-negatives $a$ and $b$ by AM-GM we obtain:
$$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.
answered yesterday
Michael Rozenberg
97.2k1589188
add a comment |
Does not exist. Try $brightarrow+infty$.
For non-negatives $a$ and $b$ by AM-GM we obtain:
$$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.
answered yesterday
Michael Rozenberg
97.2k1589188
add a comment |
Does not exist. Try $brightarrow+infty$.
For non-negatives $a$ and $b$ by AM-GM we obtain:
$$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.
answered yesterday
Michael Rozenberg
97.2k1589188
Does not exist. Try $brightarrow+infty$.
For non-negatives $a$ and $b$ by AM-GM we obtain:
$$a^2b^3=2^23^3cdotleft(frac{a}{2}right)^2left(frac{b}{3}right)^3leq2^23^3left(frac{2cdotfrac{a}{2}+3cdotfrac{b}{3}}{5}right)^5=frac{108}{3125}.$$
The equality occurs for $frac{a}{2}=frac{b}{3},$ which says that we got a maximal value.
answered yesterday
Michael Rozenberg
97.2k1589188
edited yesterday
answered yesterday
Michael Rozenberg
97.2k1589188
answered yesterday
Michael Rozenberg
97.2k1589188
answered yesterday
Michael Rozenberg
97.2k1589188
97.2k1589188
add a comment |
add a comment |
$$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
$$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
$$f'(x) = 0 implies x in {0, 1, frac 35}$$
$$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
$$f''(0)=0, f''(1)>0, f''(frac 35)<0$$
Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$
answered yesterday
Rhys Hughes
5,0441427
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
add a comment |
$$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
$$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
$$f'(x) = 0 implies x in {0, 1, frac 35}$$
$$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
$$f''(0)=0, f''(1)>0, f''(frac 35)<0$$
Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$
answered yesterday
Rhys Hughes
5,0441427
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
add a comment |
$$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
$$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
$$f'(x) = 0 implies x in {0, 1, frac 35}$$
$$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
$$f''(0)=0, f''(1)>0, f''(frac 35)<0$$
Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$
answered yesterday
Rhys Hughes
5,0441427
$$f(x)=(1-x)^2cdot x^3=x^5-2x^4+x^3$$
$$f'(x)=5x^4 -8x^3 + 3x^2=x^2(5x^2-8x+3)=x^2(x-1)(5x-3)$$
$$f'(x) = 0 implies x in {0, 1, frac 35}$$
$$f''(x)=20x^3-24x^2+6x=2x(10x^2-12x+3)$$
$$f''(0)=0, f''(1)>0, f''(frac 35)<0$$
Hence $b=frac 35, a=frac 25$, if we assume $a,b > 0$
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
answered yesterday
Rhys Hughes
5,0441427
5,0441427
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
add a comment |
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
I think it is easier to differentiate $f(x)=x^3(1-x)^2$ to $f'(x)=3x^2(1-x)^2-2x^3(1-x)$ from which it is obvious that you can take a factor $x^2(1-x)$ leaving $(3-5x)$. Also since $f(0)=f(1)=0$ and $f(x)$ is positive in $(0,1)$ there will be a local maximum in this interval. No need to take the second derivative. But neat all the same.
– Mark Bennet
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
We are searching for the absolute maximum in $[0,1]$, not local ones. As such, you need to compare the functional values at $0,1,3/5;f''(1)>0,f''(0)=0$ are insufficient to reject $0,1$
– Shubham Johri
yesterday
add a comment |
Let Constraint function be
$$C(a,b)= a+b-1=0 $$
and the Object function that needs maximization be
$$G(a,b)=a^2b^3 $$
we have with partial differentiation of combined Lagrangian:
$$ C(a,b)- lambda G(a,b) $$
condition to evaluate the multiplier $lambda$
$$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
$$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$
That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.
After plugging in these values Object function maximum value is:
$$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$
answered yesterday
Narasimham
20.6k52158
add a comment |
Let Constraint function be
$$C(a,b)= a+b-1=0 $$
and the Object function that needs maximization be
$$G(a,b)=a^2b^3 $$
we have with partial differentiation of combined Lagrangian:
$$ C(a,b)- lambda G(a,b) $$
condition to evaluate the multiplier $lambda$
$$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
$$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$
That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.
After plugging in these values Object function maximum value is:
$$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$
answered yesterday
Narasimham
20.6k52158
add a comment |
Let Constraint function be
$$C(a,b)= a+b-1=0 $$
and the Object function that needs maximization be
$$G(a,b)=a^2b^3 $$
we have with partial differentiation of combined Lagrangian:
$$ C(a,b)- lambda G(a,b) $$
condition to evaluate the multiplier $lambda$
$$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
$$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$
That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.
After plugging in these values Object function maximum value is:
$$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$
answered yesterday
Narasimham
20.6k52158
Let Constraint function be
$$C(a,b)= a+b-1=0 $$
and the Object function that needs maximization be
$$G(a,b)=a^2b^3 $$
we have with partial differentiation of combined Lagrangian:
$$ C(a,b)- lambda G(a,b) $$
condition to evaluate the multiplier $lambda$
$$dfrac {dfrac{partial C(a,b)}{partial a} } {dfrac{partial C(a,b)}{partial b} } =dfrac {dfrac{partial G(a,b)}{partial a}} {dfrac{partial G(a,b)}{partial b} } $$
$$dfrac{2a^2b^3}{3a^3b^2} =1quad rightarrow dfrac{a}{b} = dfrac{2}{3} $$
That it attains a maximum.. can be checked with second derivatives of Object function and its sign change.
After plugging in these values Object function maximum value is:
$$ dfrac{27a^5}{8} =dfrac{4b^5}{9}. $$
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
answered yesterday
Narasimham
20.6k52158
20.6k52158
add a comment |
add a comment |
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I presume you want $a$, $bge0$?
– Lord Shark the Unknown
yesterday