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Calculus 3 : Absolute Minimums and Maximums

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Example Question #1 :Applications Of Partial Derivatives

Find the absolute minimums and maximums of f(x,y)=10x^2-3y^2+10y on the disk of radius, $x^2+y^2\leq16$ .

Possible Answers:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,4)

Absolute Maximum: $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Correct answer:

Absolute Minimum: (0,-4)

Absolute Maximum: $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ , $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$

Explanation:

The first thing we need to do is find the partial derivative in respect to, and.

$\frac{\partial}{\partial x}=20x$ , $\frac{\partial}{\partial y}=-6y+10$

We need to find the critical points, so we set each of the partials equal to.

$0=20x\rightarrow x=0$

$0=-6y+10 \rightarrow y=\frac{5}{3}$

We only have one critical point at $f\Big(0, \frac{5}{3}\Big)$ , now we need to find the function value in order to see if it is inside or outside the disk.

$f\Big(0, \frac{5}{3}\Big)=10(0)^2-3\Big(\frac{5}{3}\Big)^2+10\Big(\frac{5}{3}\Big)=\frac{25}{3}\approx8.33333$

This is within our disk.

We now need to take a look at the boundary, x^2+y^2=16 . We can solve for x^2 , and plug it into.

x^2+y^2=16

x^2=16-y^2

g(y)=10(16-y^2)-3y^2+10y=-13y^2+10y+160

We will need to find the absolute extrema of this function on the range $-4\leq y \leq 4$ . We need to find the critical points of this function.

g'(y)=-26y+10

$0=-26y+10 \rightarrow y=\frac{5}{13}$

The function value at the critical points and end points are:

g(-4)=-13(-4)^2+10(-4)+160=-88

g(4)=-13(4)^2+10(4)+160=-8

$g\Big(\frac{5}{13}\Big)=-13\Big(\frac{5}{13}\Big)^2+10\Big(\frac{5}{13}\Big)+160=\frac{2105}{13}\approx161.92$

Now we need to figure out the values ofthese correspond to.

$y=-4: x^2=(-4)^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=4: x^2=4^2-16 \rightarrow x^2=0\rightarrow x=0$

$y=\frac{5}{13}: x^2=\Big(\frac{5}{13}\Big)^2-16 \rightarrow x^2=\frac{2679}{169}\rightarrow x\approx\pm\ 3.98$

Now lets summarize our results as follows:

$g(-4)=-88 \rightarrow f(0,-4)=-88$

$g(4)=-8 \rightarrow f(0,4)=-8$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(-\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

$g\Big(\frac{5}{13}\Big)=\frac{2105}{13} \rightarrow f\Big(\sqrt{\frac{2679}{169}},\frac{5}{13}\Big)=\frac{2105}{13}$

From this we can conclude that there is an absolute minimum at (0,-4) 和两个绝对最大值 $\Big(-\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ and $\Big(\sqrt{\frac{2679}{169}}, \frac{5}{13}\Big)$ .

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Calculus 3 : Absolute Minimums and Maximums

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