Optimization - rweber.net

As with related rates, optimization is sometimes approached with some dread. Fortunately, although the calculus has more steps than in related rates, the process is somewhat more cut and dried.

A. Make two-word phrase of goal, where first word is minimize or maximize and second word is the name of a quantity.
B. Name quantities and state all relationships they have, as equations. Graphs or other pictures can be helpful.
C. Make the quantity to be optimized a function of a single variable, using other relationship(s) as needed to simplify.
D. Find critical points and extrema for the function from C, and pick out the point(s) at which the goal from A is met.
E. Unpack the information from D to answer the question as stated.

Example 1.
Of all possible pairs of numbers that sum to 100, find those with the largest product.
A. Goal: maximize product.
B. Call the product and the numbers P, x, and y, respectively. We have $P = xy$ by definition and $x+y = 100$ from the problem statement.
C. Use the sum equation to simplify the product equation: $P = (100-x)\cdot x = 100x - x^2$ .
D. $P'(x) = 100 - 2x$ . CPs: $100 - 2x = 0$ only when $x = 50$ . Is this a maximum? $P'(49) = 100 - 98 = 2$ , $P'(51) = 100-102 = -2$ , so the first derivative test says yes.
E. The pair of numbers with maximum product out of all those that sum to 100 is 50 and 50.

Example 2.
A cylindrical drum must hold 10 m³ of sand. The material for the sides costs $10 per m² and the material for the base and lid cost $15 per m². Find the dimensions of the drum which minimize its cost.
A. Goal: minimize cost.
B. The volume of a cylinder is $\pi r^2 h$ , which we know equals 10. The cost of the drum is 10*(side area) + 15*(top and bottom area), which we may also put in terms of radius r and height h. Let cost be C; then $C = 10(2\pi r h) + 15(2)(\pi r^2)$ .
C. To avoid square roots, put everything in terms of r. We have $h = \frac{10}{\pi r^2}$ from the volume constraint, which means cost as a function of radius is $C(r) = 10(2\pi r \frac{10}{\pi r^2}) + 30\pi r^2 = \frac{200}{r} + 30\pi r^2$ .
D. $C'(r) = -\frac{200}{r^2} + 60\pi r$ . Critical points occur when $60\pi r = \frac{200}{r^2}$ ; solving for r we obtain $r^3 = \frac{200}{60\pi} = \frac{10}{3\pi}$ , so $r = \sqrt[3]{10/(3\pi)}$ . To find whether this is a minimum we’ll do the second derivative test. $C''(r) = 400r^{-3} + 60\pi$ . For positive r (such as our critical point) $C''$ is positive, so the graph is concave up and we have indeed found a minimum.
E. We must have a drum with a radius of $\sqrt[3]{10/(3\pi)}$ m and a height of $h = 10/\pi \cdot (10/(3\pi))^{-2/3}$ m, which simplifies to $\sqrt[3]{90/\pi}$ m. In decimal form, $r = 1.02$ m and $h = 3.06$ m.

Example 3.
Find the dimensions of the largest-area rectangle that can be inscribed in a circle of positive radius r.
Note: inscribed means the corners of the rectangle are on the circle.
A. Goal: maximize area.
B. For simplicity, let us suppose the circle is centered at the origin. Any inscribed rectangle must have a corner in each quadrant, and any corner tells us the location of the other three, by symmetry. Call the location of the first quadrant corner $(x,y)$ ; the lengths of the sides of the rectangle are $2x$ and $2y$ . Now we can wrote relationship equations: area $A = 4xy$ . Since the corners are on the circle, the distance from the origin to $(x,y)$ must be r: $r = \sqrt{x^2 + y^2}$ .
C. We must put the area equation in terms of one variable since it is what we are trying to maximize. Not having a numerical value for the radius can make it confusing, but r is a parameter: a stand-in for a constant we don’t wish to actually specify. It does not matter which variable we substitute for since we have perfect symmetry, so solve for $x = \sqrt{r^2 - y^2}$ . Then $A(y) = 4y\sqrt{r^2-y^2}$ , where r is an unspecified constant.
D. $A'(y) = 4y(1/2(r^2-y^2)^{-1/2}(-2y)) + 4\sqrt{r^2-y^2} = \frac{-4y^2}{\sqrt{r^2-y^2}} + 4\sqrt{r^2-y^2}$ . Setting this equal to zero, we obtain $4\sqrt{r^2-y^2} = \frac{4y^2}{\sqrt{r^2-y^2}}$ , which simplifies to $r^2-y^2 = y^2$ and finally $y = r/\sqrt{2}$ . Is this a maximum? Use the first derivative test. $\sqrt{2}$ is approximately 1.4; $latex 5/4 = 1.25 < \sqrt{2} < 2$, so test $latex r/2 < r/\sqrt{2} < 4r/5$. $latex A'(r/2) = \frac{-4(r^2/2)}{\sqrt{r^2 - r^2/4}} + 4\sqrt{r^2-r^2/4} = \frac{-r^2}{\sqrt{3r^2/4}} + 4\sqrt{3r^2/4} = \frac{-r^2}{\sqrt{3}/2 r} + 2\sqrt3 r$ $latex = 2\sqrt 3 r - 2r/\sqrt 3$. $latex 2\sqrt 3 > 2/\sqrt 3$ so this is positive.
$A'(4r/5) = \frac{-4(\frac{16 r^2}{25})}{\sqrt{9/25 r^2}} + 4\sqrt{9/25 r^2} = 12/5 r - 64/15 r = (36/15 - 64/15) r$ , which is negative, and hence $y = r/\sqrt 2$ is a maximum.
E. When $y = r/\sqrt 2$ , so does x, by the equation relating both of them to r. Hence a square of side length $2x = r\sqrt 2$ is the largest possible inscribed rectangle.

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