Maximize the following equation under a constraint
$begingroup$
Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
multivariable-calculus
edited Jan 22 at 0:58
David G. Stork
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
$endgroup$
add a comment |
$begingroup$
Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
multivariable-calculus
edited Jan 22 at 0:58
David G. Stork
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
$endgroup$
add a comment |
$begingroup$
Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
multivariable-calculus
edited Jan 22 at 0:58
David G. Stork
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
$endgroup$
Maximize $g(x,y)=x^4+y^4$ on $x^2+y^2=9$.
Our professor had sped through it and I didn't fully understand how they arrived at the answer. Some clarification on it would be greatly appreciated. I have attached the writing below:
multivariable-calculus
multivariable-calculus
edited Jan 22 at 0:58
David G. Stork
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
edited Jan 22 at 0:58
David G. Stork
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
edited Jan 22 at 0:58
David G. Stork
11k41432
edited Jan 22 at 0:58
David G. Stork
11k41432
edited Jan 22 at 0:58
David G. Stork
11k41432
11k41432
asked Jan 21 at 23:25
Random StudentRandom Student
443
asked Jan 21 at 23:25
Random StudentRandom Student
443
asked Jan 21 at 23:25
Random StudentRandom Student
443
443
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$
$${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$
Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:
$$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$
Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)
It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.
By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
add a comment |
$begingroup$
Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
&=&r^4(cos^4theta+sin^4theta)\
frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
&=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
&=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
&=&-2r^4sin(2theta)cos(2theta)\
&=&-r^4sin(4theta)=0
end{eqnarray}
So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.
You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.
$$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$
$$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$
$${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$
Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:
$$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$
Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)
It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.
By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
add a comment |
$begingroup$
$g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$
$${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$
Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:
$$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$
Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)
It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.
By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
add a comment |
$begingroup$
$g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$
$${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$
Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:
$$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$
Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)
It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.
By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
$endgroup$
$g(x,y) = x^4 + y^4 = x^4 + underbrace{big(sqrt{9-x^2}big)^4}_{from~constraint} = g(x)$
$${d g(x) over d x} = 4 x^3-4 x left(9-x^2right) = 4x left( x^2 - left(9-x^2right) right) = 4 x^2 (2 x^2 - 9)$$
Set this function of a single variable $x$ to equal $0$ (for an extremum) and thus find:
$$left{{xto 0},left{xto -frac{3}{sqrt{2}}right},left{xto frac{3}{sqrt{2}}right}right}$$
Then test each individually to see if it is a maximum, minimum or inflection point. (You can also calculate second-order derivatives to solve that problem, or plot the functions...)
It is clear that $x=0$ corresponds to a local maximum, while the other solutions correspond to local minima.
By the way, the handwritten notes say "obviously $(0,0)$ is a minimum," which is false. That point does not obey the constraints and is hence invalid.
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
edited Jan 22 at 1:18
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
answered Jan 21 at 23:32
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
add a comment |
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
I'm sorry, I still can't quite understand, the values that you received are the max on the boundary of the graph or just random points on the graph? I'm quite new at this.
$endgroup$
– Random Student
Jan 21 at 23:51
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
This approach misses the local extrema at $y=0$.
$endgroup$
– random
Jan 22 at 12:46
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
@random: Huh? When $y=0$, $x=pm 3$.
$endgroup$
– David G. Stork
Jan 22 at 17:33
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
$begingroup$
From the symmetries in the original problem it follows that the local maximum of $g(x,y)$ at $(0,3)$ implies the existence of the same local maximum at $(3,0)$ and $(-3,0)$. The chosen parameterization of the upper half of the circle does show them by the value of g(x) at $x=-3$ and $x=3$, but the value of $g'(x)$ at those points is misleading.
$endgroup$
– random
Jan 23 at 0:39
add a comment |
$begingroup$
Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
&=&r^4(cos^4theta+sin^4theta)\
frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
&=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
&=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
&=&-2r^4sin(2theta)cos(2theta)\
&=&-r^4sin(4theta)=0
end{eqnarray}
So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.
You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.
$$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$
$$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
add a comment |
$begingroup$
Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
&=&r^4(cos^4theta+sin^4theta)\
frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
&=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
&=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
&=&-2r^4sin(2theta)cos(2theta)\
&=&-r^4sin(4theta)=0
end{eqnarray}
So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.
You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.
$$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$
$$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
add a comment |
$begingroup$
Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
&=&r^4(cos^4theta+sin^4theta)\
frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
&=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
&=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
&=&-2r^4sin(2theta)cos(2theta)\
&=&-r^4sin(4theta)=0
end{eqnarray}
So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.
You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.
$$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$
$$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
Since we are finding a maximum on a circle centered at the origin it makes sense to convert both equations to polar coordinates:
begin{eqnarray} g(r,theta)&=&(rcostheta)^4+(rsintheta)^4\
&=&r^4(cos^4theta+sin^4theta)\
frac{dg}{dr}&=&4r^3(cos^4theta+sin^4theta)=0\
frac{dg}{dtheta}&=&-4r^4cos^3thetasintheta+4r^4sin^3thetacostheta\
&=&-4r^4sinthetacostheta(cos^2theta-sin^2theta)\
&=&-2r^4(2sinthetacostheta)(cos^2theta-sin^2theta)\
&=&-2r^4sin(2theta)cos(2theta)\
&=&-r^4sin(4theta)=0
end{eqnarray}
So extrema should occur when $sin(4theta)=0$. And $sin(4theta)=0$ only when $4theta$ is a whole multiple of $pi$. Thus you must check points of the circle corresponding to multiples of $dfrac{pi}{4}$.
You should find that the minima occur when $theta$ is an odd multiple of $dfrac{pi}{4}$ and the maxima will occur when $theta$ is an even multiple of $dfrac{pi}{4}$ (i.e. a multiple of $dfrac{pi}{2})$.
$$gleft(3,frac{pi}{4}right)=gleft(3,frac{3pi}{4}right)=gleft(3,frac{5pi}{4}right)=gleft(3,frac{7pi}{4}right)=left(pmfrac{3}{sqrt{2}}right)^4+left(pmfrac{3}{sqrt{2}}right)^4=frac{81}{2}$$
$$g(3,0)=gleft(3,frac{pi}{2}right)=gleft(3,piright)=gleft(3,frac{3pi}{2}right)=3^4=81$$
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
edited Jan 26 at 20:37
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
answered Jan 22 at 0:15
John Wayland BalesJohn Wayland Bales
14.5k21238
14.5k21238
add a comment |
add a comment |
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