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

24 min Optimization
Applied max-min problems translate real-world scenarios—maximizing area, minimizing cost, optimizing dimensions—into calculus optimization problems. The typical workflow is: (1) identify the quantity to optimize, (2) express it as a function of one variable using constraint equations, (3) find critical points via differentiation, and (4) verify the answer using endpoint checks or the second derivative test. Classic examples include maximizing the volume of a box with fixed surface area, finding the shortest ladder over a wall, and minimizing travel time across different media (Snell's law). The key challenge is often the algebraic setup—correctly eliminating variables and identifying the feasible domain.

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.

  1. Let $x$ = side perpendicular to barn. Fence: $2x + y = 200$, so $y = 200 - 2x$.
  2. $A = xy = x(200-2x) = 200x - 2x^2$.
  3. $A' = 200 - 4x = 0 \Rightarrow x = 50$.
  4. $y = 100$.
  5. $A_{\max} = 5{,}000\text{ m}^2$.
Example 2

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

  1. $S = x^2 + 4xh = 48 \Rightarrow h = (48-x^2)/(4x)$.
  2. $V = x^2h = x(48-x^2)/4 = 12x - x^3/4$.
  3. $V' = 12 - 3x^2/4 = 0 \Rightarrow x = 4$.
  4. $h = 2$. $V_{\max} = 32\text{ cm}^3$.
Example 3

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

  1. $D^2 = x^2 + (x^2-4)^2 = x^4 - 7x^2 + 16$.
  2. $\frac{d}{dx}(D^2) = 4x^3 - 14x = 2x(2x^2-7) = 0$.
  3. $x = 0$ or $x = \pm\sqrt{7/2}$.
  4. At $x = \sqrt{7/2}$: $D^2 = 7/2 + (7/2-4)^2 = 7/2+1/4 = 15/4$.
  5. $D = \sqrt{15}/2$.

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 $= r/\sqrt{2}$; $A = r^2$.

📦 Box Volume Maximizer (fixed surface area)
Optimal side length
Max volume (cube)
Sphere of same S: volume
Sphere advantage
Unconstrained Optimization Linear Programming