# 13.9E: Optimization of Functions of Several Variables (Exercises)

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## Absolute Extrema on Closed and Bounded Regions

In exercises 1 - 4, find the absolute extrema of the given function on the indicated closed and bounded set $$R$$.

1) $$f(x,y)=xy−x−3y; R$$ is the triangular region with vertices $$(0,0),(0,4),$$ and $$(5,0)$$.

2) Find the absolute maximum and minimum values of $$f(x,y)=x^2+y^2−2y+1$$ on the region $$R=\{(x,y)∣x^2+y^2≤4\}.$$

$$f$$ has an absolute minimum of $$0$$ at $$(0, 1)$$ and an absolute maximum of $$9$$ at $$(0,−2)$$.

3) $$f(x,y)=x^3−3xy−y^3$$ on $$R=\{(x,y):−2≤x≤2,−2≤y≤2\}$$

4) $$f(x,y)=\frac{−2y}{x^2+y^2+1}$$ on $$R=\{(x,y):x^2+y^2≤4\}$$

$$f$$ has an absolute minimum of $$-1$$ at $$(0,1)$$ and an absolute maximum of $$1$$ at $$(0,−1)$$.

5)  To be created...

6) Find the absolute minimum and absolute maximum of the function $$f(x,y) = xy^2 - 2x + 3$$ on the region between $$y = x/2$$ and $$y = \sqrt{x}$$, for $$0 \le x \le 4$$, showing all work, including work to find any critical points of $$f$$.  See the plot of the region in the $$xy$$-plane below.

$$f$$ has an absolute minimum of   $$3 - \frac{8\sqrt{6}}{9} \approx 0.822676$$   at $$\left( \frac{2\sqrt{6}}{3}, \frac{\sqrt{6}}{3} \right)$$ and an absolute maximum of $$11$$ at $$(4, 2)$$.
Solution:
First, we find the critical points of $$f$$.  The partials of $$f$$ are:
\begin{align*} f_x(x,y) &= y^2 - 2 \\ f_y(x,y) &= 2xy \end{align*}
Setting the partials both equal to $$0$$, we get the equations:
$y^2 - 2 = 0 \qquad \text{and} \qquad 2xy = 0 \nonumber$
Solving the first for $$y$$ gives us: $$y = \pm \sqrt{2}$$.

If $$y = \sqrt{2}$$, the second equation becomes $$2x(\sqrt{2}) = 0$$.  Solving this equation for $$x$$, we get $$x = 0$$.

Note that if $$y = -\sqrt{2}$$, we also get $$x = 0$$.

Thus, the critical points of $$f$$ are: $$(0, -\sqrt{2})$$ and $$(0, \sqrt{2})$$.

Since neither of these critical points is inside the closed, bounded region we are looking at in this problem, we can reject them as locations of absolute extrema of $$f$$ on this region.

Now let's consider the two boundary curves, remembering to check the endpoints of the interval as we do this.

If $$y = \sqrt{x}$$, we get the function of one variable, \begin{align*} g(x) = f(x, \sqrt{x}) &= x (\sqrt{x})^2 - 2x + 3 \\ &= x^2 - 2x + 3 \end{align*}
Now let's locate any critical values for this function $$g$$.
$\text{Set }g'(x) = 2x -2 =0. \nonumber$
Then $$x = 1$$ is a critical value of $$g$$.

Since $$x = 1$$ is in the interval $$[0, 4]$$, we evaluate $$g$$ at this critical value and at the endpoints of the interval, $$x = 0$$ and $$x = 4$$.
\begin{align*} g(1) &= f(1, 1) = 1(1)^2 - 2(1) + 3 = \boxed{2} \\[5pt] g(0) &= f(0, 0) = 0(0)^2 - 2(0) + 3 =\boxed{ 3 }\\[5pt] g(4) &= f(4, 2) = 4(2)^2 -2(4) + 3 = \boxed{ 11 } \end{align*}
Now, if $$y = x/2$$, we get the function of one variable, \begin{align*} h(x) = f(x, \tfrac{x}{2}) &= x \left(\tfrac{x}{2}\right)^2 - 2x + 3 \\ &= \frac{x^3}{4} - 2x + 3 \end{align*}
Locating any critical values for this function $$h$$, we
$\text{Set }h'(x) = \frac{3x^2}{4} -2 =0. \nonumber$
Solving this equation for $$x$$, we solve as follows. \begin{align*} 3x^2 &= 8 \\[5pt] x^2 &= \frac{8}{3}\\[5pt] x &= \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{6}}{3} \end{align*}
Thus $$h$$ has two critical values:  $$x = -\frac{2\sqrt{6}}{3}$$ and $$x = \frac{2\sqrt{6}}{3} \approx 1.633$$.

Since the negative critical number is not in the interval $$[0, 4]$$, we can reject it.  But the positive critical number is in this interval, so we evaluate $$h$$ at this value.
\begin{align*} h\left(\frac{2\sqrt{6}}{3}\right) &= f\left(\frac{2\sqrt{6}}{3}, \frac{\sqrt{6}}{3} \right) = \frac{2\sqrt{6}}{3}\left(\frac{\sqrt{6}}{3}\right)^2 - 2\left(\frac{2\sqrt{6}}{3}\right) + 3 \\[5pt] &= \frac{4\sqrt{6}}{9} - \frac{4\sqrt{6}}{3} + 3 = \boxed{3 - \frac{8\sqrt{6}}{9}} \approx 0.822676 \\[5pt] h(0) &= f(0, 0) = 0(0)^2 - 2(0) + 3 = \boxed{ 3 } \\[5pt] h(4) &= f(4, 2) = 4(2)^2 -2(4) + 3 = \boxed{ 11 } \end{align*}
Comparing the values of $$f$$ at all these points, we find that:

$$f$$ has an absolute minimum of   $$3 - \frac{8\sqrt{6}}{9} \approx 0.822676$$   at $$\left( \frac{2\sqrt{6}}{3}, \frac{\sqrt{6}}{3} \right)$$ and an absolute maximum of $$11$$ at $$(4, 2)$$.

## Applications

7) Find three positive numbers the sum of which is $$27$$, such that the sum of their squares is as small as possible.

8) Find the points on the surface $$x^2−yz=5$$ that are closest to the origin.

Hint:
Use the distance formula. But note that you can leave off the square root, since the minimum value of the square of the distance will also minimize the distance.
$$(\sqrt{5},0,0),(−\sqrt{5},0,0)$$

9) Find the maximum volume of a rectangular box with three faces in the coordinate planes and a vertex in the first octant on the plane $$x+y+z=1$$.

10) The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed $$108$$ in. Find the dimensions of the rectangular package of largest volume that can be sent.

$$18$$ in. by $$36$$ in. by $$18$$ in.

11) A cardboard box without a lid is to be made with a volume of $$4$$ ft3. Find the dimensions of the box that requires the least amount of cardboard.

12) Find the point on the surface $$f(x,y)=x^2+y^2+10$$ nearest the plane $$x+2y−z=0.$$ Identify the point on the plane.

Hint:
Here one approach is the find the distance between a point $$(x_0, y_0, z_0)$$ on the surface and the plane, using what you learned in Section 11.5. Then you can substitute the surface function into this distance function for $$z_0$$ and substitute $$x$$ for $$x_0$$ and $$y$$ for $$y_0$$. This will give you a function of $$x$$ and $$y$$ that you can minimize.
$$(0.5,1,11.25)$$ is the point on the surface nearest the plane.
Although it was not requested, note that $$(\frac{47}{24},\frac{47}{12},\frac{235}{24})$$ is the point on the plane that is nearest the surface.
See this problem illustrated in CalcPlot3D.

13) Find the point in the plane $$2x−y+2z=16$$ that is closest to the origin.

14) A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from $$x$$ units of jogging shoes and $$y$$ units of cross-trainers is given by $$R(x,y)=−5x^2−8y^2−2xy+42x+102y,$$ where $$x$$ and $$y$$ are in thousands of units. Find the values of $$x$$ and $$y$$ to maximize the total revenue.

$$x=3$$ and $$y=6$$

15) A shipping company handles rectangular boxes provided the sum of the length, width, and height of the box does not exceed $$96$$in. Find the dimensions of the box that meets this condition and has the largest volume.

16) Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is $$120$$ cm.

$$V=\frac{64,000}{π}\,\text{cm}^3≈20,372\,\text{cm}^3$$