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emwhy's question at Yahoo! Answers regarding finding extrema in a given region
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Re: emwhy's question at Yahoo! Questions regarding finding extrema in a given region
Hello emwhy,
First, let's find the critical values inside the given region. We do this by equating the partials to zero, and solving the resulting system:
We are given:
\(\displaystyle f(x,y)=6x^2+3y^2-36x-1\)
Hence:
\(\displaystyle f_x(x,y)=12x-36=12(x-3)=0\)
\(\displaystyle f_y(x,y)=6y=0\)
This gives us the critical point $(x,y)=(3,0)$. Since its distance from the origin is less than $6$, we know it is within the circular region.
Next, we may employ the second partials test to determine the nature of the critical point.
Let:
\(\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y) \right)^2\)
Thus:
\(\displaystyle D(x,y)=12\cdot6-\left(0 \right)^2=72>0\)
Since \(\displaystyle f_{xx}(3,0)=12>0\) we conclude that the critical point $(3,0)$ is a relative minimum.
Now we want to check the boundary. In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=6^2$ by means of the parametric equations $x=6\cos(t),\,y=6\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:
\(\displaystyle f(\cos(t),\sin(t))=f(t)=6\left(6\cos(t) \right)^2+3\left(6\sin(t) \right)^2-36\left(6\cos(t) \right)-1=108\cos^2(t)-216\cos(t)+107\)
Differentiating with respect to $t$ and equating to zero, we find:
\(\displaystyle f'(t)=-216\cos(t)\sin(t)+216\sin(t)=216\sin(t)\left(1-\cos(t) \right)=0\)
This gives us the critical values for $t$:
\(\displaystyle t=0,\,\pi,\,2\pi\)
Since the first and third critical values are boundary values for $t$, and because the parametrization of $x$ and $y$ have the same value for these two values, we need only consider:
\(\displaystyle t=0,\,\pi\)
This gives us the two critical points:
\(\displaystyle \left(x(0),y(0) \right)=(6,0)\)
\(\displaystyle \left(x(\pi),y(\pi) \right)=(-6,0)\)
Now, we may use the second derivative test to determine the nature of these extrema. Let's write the first derivative as:
\(\displaystyle f'(t)=-108\sin(2t)+216\sin(t)\)
Differentiating, we find:
\(\displaystyle f''(t)=-216\cos(2t)+216\cos(t)\)
Checking the critical values:
\(\displaystyle f''(0)=0\)
The second derivative test is inconclusive for this point, so using the first derivative test, we find:
\(\displaystyle f'\left(\frac{\pi}{2} \right)=216>0\)
\(\displaystyle f'\left(\frac{3\pi}{2} \right)=-216<0\)
Hence the point $(6,0)$ is a relative minimum.
\(\displaystyle f''(\pi)=-432<0\)
Hence the point $(-6,0)$ is a relative maximum.
So, we now know:
\(\displaystyle f_{\max}=f(-6,0)=6(-6)^2+3(0)^2-36(-6)-1=431\)
\(\displaystyle f_{\min}=f(3,0)=6(3)^2+3(0)^2-36(3)-1=-55\)
Hello emwhy,
First, let's find the critical values inside the given region. We do this by equating the partials to zero, and solving the resulting system:
We are given:
\(\displaystyle f(x,y)=6x^2+3y^2-36x-1\)
Hence:
\(\displaystyle f_x(x,y)=12x-36=12(x-3)=0\)
\(\displaystyle f_y(x,y)=6y=0\)
This gives us the critical point $(x,y)=(3,0)$. Since its distance from the origin is less than $6$, we know it is within the circular region.
Next, we may employ the second partials test to determine the nature of the critical point.
Let:
\(\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y) \right)^2\)
Thus:
\(\displaystyle D(x,y)=12\cdot6-\left(0 \right)^2=72>0\)
Since \(\displaystyle f_{xx}(3,0)=12>0\) we conclude that the critical point $(3,0)$ is a relative minimum.
Now we want to check the boundary. In order to examine $f$ on the boundary of the region, we may represent the circle $x^2+y^2=6^2$ by means of the parametric equations $x=6\cos(t),\,y=6\sin(t),\,0\le t\le 2\pi$. Thus, on the boundary we can write $f$ as a function of a single variable $t$:
\(\displaystyle f(\cos(t),\sin(t))=f(t)=6\left(6\cos(t) \right)^2+3\left(6\sin(t) \right)^2-36\left(6\cos(t) \right)-1=108\cos^2(t)-216\cos(t)+107\)
Differentiating with respect to $t$ and equating to zero, we find:
\(\displaystyle f'(t)=-216\cos(t)\sin(t)+216\sin(t)=216\sin(t)\left(1-\cos(t) \right)=0\)
This gives us the critical values for $t$:
\(\displaystyle t=0,\,\pi,\,2\pi\)
Since the first and third critical values are boundary values for $t$, and because the parametrization of $x$ and $y$ have the same value for these two values, we need only consider:
\(\displaystyle t=0,\,\pi\)
This gives us the two critical points:
\(\displaystyle \left(x(0),y(0) \right)=(6,0)\)
\(\displaystyle \left(x(\pi),y(\pi) \right)=(-6,0)\)
Now, we may use the second derivative test to determine the nature of these extrema. Let's write the first derivative as:
\(\displaystyle f'(t)=-108\sin(2t)+216\sin(t)\)
Differentiating, we find:
\(\displaystyle f''(t)=-216\cos(2t)+216\cos(t)\)
Checking the critical values:
\(\displaystyle f''(0)=0\)
The second derivative test is inconclusive for this point, so using the first derivative test, we find:
\(\displaystyle f'\left(\frac{\pi}{2} \right)=216>0\)
\(\displaystyle f'\left(\frac{3\pi}{2} \right)=-216<0\)
Hence the point $(6,0)$ is a relative minimum.
\(\displaystyle f''(\pi)=-432<0\)
Hence the point $(-6,0)$ is a relative maximum.
So, we now know:
\(\displaystyle f_{\max}=f(-6,0)=6(-6)^2+3(0)^2-36(-6)-1=431\)
\(\displaystyle f_{\min}=f(3,0)=6(3)^2+3(0)^2-36(3)-1=-55\)