Training Optimization Applied Max-Min Problems
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Applied Max-Min Problems

24 min Optimization

Applied Max-Min Problems

Strategy
  1. Draw a diagram, define variables
  2. Write the quantity to optimize as a function of one variable
  3. Find the domain (physical constraints)
  4. Take the derivative, set to zero, solve
  5. Verify it's a max or min; state the answer
Example 1

A farmer has 200 m of fencing to enclose a rectangular field against a barn (no fence needed on the barn side). Maximize the area.

Let $x$ = side perpendicular to barn. Fence: $2x + y = 200$, so $y = 200 - 2x$.

$A = xy = x(200-2x) = 200x - 2x^2$.

$A' = 200 - 4x = 0 \Rightarrow x = 50$. $y = 100$. $A_{\max} = 5{,}000\text{ m}^2$.

Example 2

Find the dimensions of a box with square base and open top that has maximum volume with surface area $48\text{ cm}^2$.

$S = x^2 + 4xh = 48 \Rightarrow h = (48-x^2)/(4x)$.

$V = x^2h = x(48-x^2)/4 = 12x - x^3/4$.

$V' = 12 - 3x^2/4 = 0 \Rightarrow x = 4$. $h = 2$. $V_{\max} = 32\text{ cm}^3$.

Example 3

Minimize the distance from $(0,0)$ to the parabola $y = x^2 - 4$.

$D^2 = x^2 + (x^2-4)^2 = x^4 - 7x^2 + 16$.

$\frac{d}{dx}(D^2) = 4x^3 - 14x = 2x(2x^2-7) = 0$. $x = 0$ or $x = \pm\sqrt{7/2}$.

At $x = \sqrt{7/2}$: $D^2 = 7/2 + (7/2-4)^2 = 7/2+1/4 = 15/4$. $D = \sqrt{15}/2$.

Practice Problems

1. Maximize area of a rectangle with perimeter 40.
2. Two positive numbers sum to 20. Maximize their product.
3. A 12-inch wire is cut into a square and circle. Minimize total area.
4. Minimize cost: cylindrical can, volume $1000\text{ cm}^3$, top + bottom cost $\$2/\text{cm}^2$, side $\$1/\text{cm}^2$.
5. Shortest fence: divide a rectangular field of area 600 m² in half with a fence parallel to one side.
6. Revenue: $R = 100x - x^2$. Maximize revenue.
7. Maximize the area of a triangle with base on the $x$-axis inscribed under $y = 4 - x^2$.
8. Closest point on $y = \sqrt{x}$ to $(3, 0)$.
9. Maximum volume of a cone inscribed in a sphere of radius $R$.
10. Profit: $P = -2x^2 + 100x - 300$. Max profit?
11. Optimal speed: fuel cost per mile is $v/200$ dollars, driver pay $\$15$/hr. Minimize cost/mile.
12. Largest rectangle inside a semicircle of radius $r$.
Show Answer Key

1. Square $10 \times 10$; $A = 100$

2. $x = y = 10$; product $= 100$

3. Let $x$ = square perimeter. $A = x^2/16 + (12-x)^2/(4\pi)$. Min at $x = \frac{48}{4+\pi}$.

4. $C = 4\pi r^2 + 2\pi rh$, $V = \pi r^2 h = 1000$. Solve $C'(r)=0$.

5. $F = 2x + 3y$, $xy = 600$. Min at $x = 30$, $y = 20$; $F = 120$ m.

6. $R'=100-2x=0$; $x=50$. $R=2500$.

7. Area $= (2x)(4-x^2) = 8x - 2x^3$. Max at $x = 2/\sqrt{3}$; $A = 32\sqrt{3}/9$.

8. Min $D^2 = (x-3)^2+x$. $D^2'=2(x-3)+1=0$; $x=5/2$. Point $(5/2, \sqrt{5/2})$.

9. $V_{\max} = \frac{32\pi R^3}{81}$

10. $P'=-4x+100=0$; $x=25$. $P=\$950$.

11. Cost/mile $= v/200 + 15/v$. Min at $v = \sqrt{3000} \approx 54.8$ mph.

12. Width $= r\sqrt{2}$, height $= r/\sqrt{2}$; $A = r^2$.

Unconstrained Optimization Linear Programming