Calculus III Advanced (Course) (13.8) (Homework)

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Section 13.8 Homework

From Calculus 10e by Larson and Edwards, p. 942. Exercises 6, 12, 18, 27, 28, 29, 30, 42, 46

Exercise 13.8.6 Finding Relative Extrema

Identify an extrema for the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points and test for relative extrema.

\( f(x,y) = -x^{2}-y^{2}+10x+12y -64 \)

Solution Wolfram Alphfa[1] Find the critical points

\(f_{x}(x,y) = -2x+10 \)
\(f_{y}(x,y) = -2y+12 \)

\(-2x+10 = 0 \rightarrow x=5\) and \(-2y+12 = 0 \rightarrow y = 6\) yields the critical point \((5,6)\). Completing the square for \(f\) yields all \((x,y) \ne (5,6)\)

\(max: f(x,y) = -(x-5)^{2}-(y-6)^{2}+10(x-5)+12(y-6)-64 =-3\)

The maximum is \((5,6,-3)\), as shown in Figure 1.

Figure 1

Exercise 13.8.12 Using the Second Partials Test

Examine the function for relative extrema and saddle points.

\( f(x,y) = 2x^{2}+2xy+y^{2}+2x-3\)

Solution Wolfram Alphfa[2]
The critical point is \((-1,1,-4)\) which is a minimum, there are no addle points, as shown in Figure 2.

Figure 2

Exercise 13.8.18 Using the Second Partials Test

Examine the function for relative extrema and saddle points.

\(f(x,y) = 2xy-1/2(x^{4}+y^{4}) + 1 \)

Solution Find the critical points.

\(f_{x}(x,y) = 2y-2x^{3}\)
\(f_{y}(x,y) = 2x-2y^{3}\)
\(f_{xx}(x,y) = -6x^{2}\)
\(f_{yy}(x,y) = -6y^{2}\)
\(f_{xy}(x,y) = 2\)
\( x = 0, \pm 1\) and \(y = \pm 1 \)

Relative maximum points are \((1,1,2)\) and \((-1,-1,2)\).

\(d\) \(=(-6x^{2})(-6y^{2}) - [2]^{2} \)
\(= (-6x^{2})(-6y^{2}) - 4 \)
\(= -4 \)

\((0,0)\) is the saddle point.

Figure 3

Exercise 13.8.27 Think About It

Determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information for the function.

\( f_{xx}(x_{0}, y_{0})=9, \: f_{yy}(x_{0}, y_{0})=4, \: f_{xy}(x_{0}, y_{0})=6 \)

Solution

\( d= (9)(4)-(6)^2 = 0\)

Insufficient information.

Exercise 13.8.28 Think About It

Determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information for the function.

\( f_{xx}(x_{0}, y_{0})=-3, \: f_{yy}(x_{0}, y_{0})=-8, \: f_{xy}(x_{0}, y_{0})=2 \)

Solution

\( d= (-3)(-8)-(2)^2 = 22\)

\(d > 0 \) and \(f_{xx}(x_{0}, y_{0}) < 0\), \(f\) has a relative maximum at \((a,b)\) and no saddle point.

Exercise 13.8.29 Think About It

Determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information for the function.

\( f_{xx}(x_{0}, y_{0})=-9, \: f_{yy}(x_{0}, y_{0})=6, \: f_{xy}(x_{0}, y_{0})=10 \)

Solution

\( d= (-9)(6)-(10)^2 =-54-100 = -154\)

\(d < 0 \) and has saddle point, but no relative max or min.

Exercise 13.8.30 Think About It

Determine whether there is a relative maximum, a relative minimum, a saddle point, or insufficient information for the function.

\( f_{xx}(x_{0}, y_{0})=25, \: f_{yy}(x_{0}, y_{0})=8, \: f_{xy}(x_{0}, y_{0})=10 \)

Solution

\( d= (25)(8)-(10)^2 = 200-100 = 100\)

\(d > 0\) and \( f_{xx}(x_{0}, y_{0}) > 0 \), \(f\) has a relative minimum at \((a,b)\)

Exercise 13.8.42 Finding Absolute Extrema

Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.)

\(f(x,y) = x^{2}+xy, \: R=\{(x,y): |x| \leqslant 2, \: |y| \leqslant 1 \} \)

Solution Part 1 Saddle Point or not?
Take the partial derivatives

\( f_{x}(x,y) = 2x+y =0 \) and \( f_{y}(x,y) = x = 0 \) So, \((x,y) = (0,0)\), the Second Partial Test is \(=-1 < 0 \). (0,0) is a saddle point.

Part 2 Find the extrema on the boundaries.
\(S_{1} = \{ x=2, |y| \leqslant 1 \}. \: g_{1}(y)=f(2,y) = 4+2y [-1,1].\)

$$ \frac{\partial g_{1}}{\partial y} = 2 \ne 0$$

So no critical point along the side.

The end points are \((2,-1)\) and \((2,1)\). They yield

\( f(2,-1) = g_{1}(-1) = 1, \: f(2,1) = g_{1}(1) = 6 \)

\(S_{2} = \{ y=1, |x| \leqslant 2 \}. \: g_{2}(x)=f(x,1) = x^{2}+x [-2,2].\)

$$ \frac{\partial g_{2}}{\partial x} = 2x+1.$$

There is a critical point along the side at \(x= 1/2\).

The end points are \((-1/2,1)\) and \((-2,1)\). They yield

\( f(-1/2,-1) = g_{2}(-1/2) = -1/4, \: f(-2,1) = g_{2}(-2) = 2 \)

\(S_{3} = \{x=-2, |y| \leqslant 1 \}. \: g_{3}(y)=f(-2,y) = 4-2y [-1,1].\)

$$ \frac{\partial g_{3}}{\partial y} = -2 \ne 0.$$

There is no critical point along the side.

The new end point is \((-1/2,1)\) and \((-2,-1)\). It yields

\( f(-2,-1) = g_{3}(-1) = 6 \)

\(S_{4} = \{y=-1, |x| \leqslant 2 \}. \: g_{4}(x)=f(x,-1) = x^{2}-x [-2,2].\)

$$ \frac{\partial g_{4}}{\partial x} = 2x-1.$$

There is a critical point along the side at \(x=1/2\).

It yields

\( f(1/2,-1) = g_{4}(1/2) = -1/4 \)

This produces two absolute maximum points: \(f(2,1) = f(-2,1) = 6\), and two absolute minimum points: \(f(-1/2,1)= f(1/2,-1)=-1/4\).

As shown in Figure 4.

Figure 4

Exercise 13.8.46 Finding Absolute Extrema

Find the absolute extrema of the function over the region \(R\). (In each case, R contains the boundaries.)

\(f(x,y) = 2x-2xy+y^{2} \)

\(R\) is the region in the \(xy\)-plane bounded by the graphs for \(y=x^{2}\) and \(y=1\)
Solution
1. Critical point.

\(f_{x} = 2-2y = 0, \: f_{y}=-2x+2y = 0 \rightarrow (x,y) = (1,1).\)

This is the absolute minimum.
2. The boundary has two parts.
\(C_{1} = \{ y = x^{2}, \: |x| \leqslant 1 \}. \: g_{1}(x)=f(x,x^{2})=2x-2x^{3}+x^{4}, \: [-1,1]\). \(\partial g_{1}/\partial x = 2-6x^{2} + 4x^{3} = 2(1-3x^{2}+2x^{3})\). Factoring produces

\(= (x-1)(x-1)(2x+1). \)

This yields two critical points, \(x=1, \: -1/2, (-1/2,1/4),(1,1),(-1,1)\). The values are

\(f(-1/2,1/4)=g_{1}(-1/2) = -11/16, f(1,1)=g_{1}(1) = 1,f(-1,1) = 1.\)

\(C_{2} = \{y=1, |x| \leqslant 1\}. g_{1}(x)=f(x,1) = 1 \: [-1,1].\) So \(f\) is a constant along \(C_{2}\) and every point is a candidate point.

The absolute minimum: \(f(-1/2,1/4) = -11/16\), and \(f\) obtains absolute maximum value 1 at every point on the line segment \(C_{2}\).

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Parent Article: Calculus III Advanced (Course)