Introduction

This topic focuses on solving complete optimization problems and interpreting results in context. The essential knowledge tells us that minimum and maximum values of a function take on specific meanings in applied contexts. Understanding how to translate real-world situations into mathematical optimization problems and interpret the solutions is crucial for applying calculus to practical scenarios.

Complete Optimization Strategy

Systematic Problem-Solving Approach

Step 1: Read and Understand

Step 2: Draw a Diagram

Step 3: Write Equations

Step 4: Eliminate Variables

Step 5: Optimize

Step 6: Interpret

Geometric Optimization Problems

Maximizing Area with Fixed Perimeter

Example 1:

A farmer has 400 feet of fencing to enclose a rectangular field. One side of the field is along a river and requires no fencing. What dimensions maximize the enclosed area?

Step 1-2: Need to maximize area. Let $x$ = length parallel to river, $y$ = width perpendicular to river.

Step 3:

Constraint: $x+2y=400$

Objective: Maximize $A=xy$

Step 4:

From constraint: $x=400-2y$

Substitute: $A(y)=(400-2y)y=400y-2y^2$

Domain: $0<y<200$

Step 5:

$A^{\prime }(y)=400-4y$

$400-4y=0$

$y=100$

Then $x=400-200=200$

Check: $A(100)=200(100)=20000$ square feet

Step 6: The field should be 200 feet along the river and 100 feet perpendicular to the river, creating a maximum area of 20,000 square feet. This makes sense because the optimal rectangle uses twice as much fencing for the width (two sides) as for the length (one side).

Minimizing Surface Area with Fixed Volume

Example 2:

A cylindrical can must hold 1000 cm³. Find the dimensions that minimize the amount of material (surface area) used.

Step 1-2: Minimize surface area. Let $r$ = radius, $h$ = height.

Step 3:

Constraint: $V=\pi r^{2}h=1000$

Objective: Minimize $S=2\pi r^2+2\pi rh$ (top, bottom, and side)

Step 4:

From constraint: $h=\frac{1000}{\pi r^2}$

Substitute: $S(r)=2\pi r^2+2\pi r\cdot \frac{1000}{\pi r^2}=2\pi r^2+\frac{2000}{r}$

Domain: $r>0$

Step 5:

$S^{\prime }(r)=4\pi r-\frac{2000}{r^2}$

$4\pi r-\frac{2000}{r^2}=0$

$4\pi r=\frac{2000}{r^2}$

$4\pi r^3=2000$

$r^3=\frac{500}{\pi }$

$r=\sqrt[3]{\frac{500}{\pi }}\approx 5.42$ cm

Then $h=\frac{1000}{\pi r^2}=\frac{1000}{\pi \cdot \frac{500}{\pi }}=2r\approx 10.84$ cm

Step 6: The can should have radius approximately 5.42 cm and height approximately 10.84 cm to minimize material. Interestingly, the optimal can has height equal to its diameter ($h=2r$), though real-world cans often differ due to manufacturing and marketing considerations.

Distance and Proximity Problems

Minimizing Distance to a Curve

Example 3:

Find the point on the parabola $y=x^2$ that is closest to the point $(3,0)$.

Step 1-2: Minimize distance from $(x,x^2)$ on the parabola to $(3,0)$.

Step 3: Distance: $D=\sqrt{(x-3{)}^2+(x^2-0{)}^2}$

Step 4:

To simplify, minimize $D^2$ (same critical points): $f(x)=(x-3{)}^2+(x^2{)}^2=x^2-6x+9+x^4$

Domain: all real numbers

Step 5: $f^{\prime }(x)=2x-6+4x^3$

$4x^3+2x-6=0$

$2x^3+x-3=0$

By inspection or technology: $x=1$ is a solution

$(2x^3+x-3)=(x-1)(2x^2+2x+3)=0$

The quadratic factor has no real roots, so $x=1$ is the only critical point.

Point on parabola: $(1,1)$

Step 6: The closest point on the parabola to $(3,0)$ is $(1,1)$, with a minimum distance of $\sqrt{(1-3{)}^2+1^2}=\sqrt{5}\approx 2.24$ units. This point represents where a person standing at $(3,0)$ would walk the shortest path to reach the curve.

Business and Economics Applications

Maximizing Revenue

Example 4:

A company can sell 1000 units per week at $50 per unit. For each $1 decrease in price, they sell 50 more units. What price maximizes revenue?

Step 1-2: Maximize revenue. Let $x$ = number of $1 decreases in price.

Step 3:

Price per unit: $p=50-x$

Quantity sold: $q=1000+50x$

Revenue: $R=p\cdot q$

Step 4:

$R(x)=(50-x)(1000+50x)$

$R(x)=50000+2500x-1000x-50x^2$

$R(x)=50000+1500x-50x^2$

Domain: $0\le x\le 50$ (price can't be negative)

Step 5:

$R^{\prime }(x)=1500-100x$

$1500-100x=0$

$x=15$

Price: $p=50-15=35$ dollars

Revenue: $R(15)=35(1250)=43750$ dollars

Step 6: The company should charge $35 per unit, selling 1,250 units per week for maximum weekly revenue of $43,750. This represents a balance between lower prices (which increase sales) and maintaining sufficient profit per unit.

Minimizing Average Cost

Example 5:

A company's cost function is $C(x)=1000+5x+0.01x^2$ dollars for producing $x$ units. Find the production level that minimizes average cost.

Step 1-2: Minimize average cost per unit.

Step 3:

Average cost: $\bar{C}(x)=\frac{C(x)}{x}=\frac{1000+5x+0.01x^2}{x}$

Step 4:

$\bar{C}(x)=\frac{1000}{x}+5+0.01x$

Domain: $x>0$

Step 5:

${\bar{C}}^{\prime }(x)=-\frac{1000}{x^2}+0.01$

$-\frac{1000}{x^2}+0.01=0$

$0.01=\frac{1000}{x^2}$

$0.01x^2=1000$

$x^2=100000$

$x=\sqrt{100000}\approx 316.23$

Average cost: $\bar{C}(316.23)\approx 11.16$ dollars per unit

Step 6: The company should produce approximately 316 units to minimize average cost at about $11.16 per unit. Producing fewer units spreads fixed costs over fewer items (expensive), while producing more increases variable costs per item. The optimal point balances these factors.

Physical Applications

Projectile Motion Optimization

Example 6:

A ball is thrown from ground level with initial velocity 64 ft/s. Its height is $h(t)=64t-16t^2$ feet after $t$ seconds. What is the maximum height reached?

Step 1-2: Maximize height.

Step 3: $h(t)=64t-16t^2$

Domain: $0\le t$ until ball lands

Step 4: No variables need to be eliminated, as height is expressed with respect to time

Step 5: $h^{\prime }(t)=64-32t$ $64-32t=0$ $t=2$ seconds

Maximum height: $h(2)=64(2)-16(4)=128-64=64$ feet

Step 6: The ball reaches a maximum height of 64 feet after 2 seconds. At this point, the velocity is zero (the ball momentarily stops before falling back down). This represents the apex of the parabolic trajectory.

Construction and Design Problems

Minimizing Material Cost

Example 7:

A rectangular storage container with an open top must have a volume of 10 m³. The base costs $4 per m² and the sides cost $2 per m². Find dimensions that minimize cost.

Step 1-2: Minimize total cost. Let base be $x$ by $y$ meters, height $h$ meters.

Step 3:

Constraint: $V=xyh=10$

Cost: $C=4xy+2(2xh+2yh)=4xy+4xh+4yh$

Step 4:

From constraint: $h=\frac{10}{xy}$

Substitute: $C(x,y)=4xy+4x\cdot \frac{10}{xy}+4y\cdot \frac{10}{xy}=4xy+\frac{40}{y}+\frac{40}{x}$

For simplicity, assume square base ($x=y$): $C(x)=4x^2+\frac{80}{x}$

Domain: $x>0$

Step 5: $C^{\prime }(x)=8x-\frac{80}{x^2}$

$8x-\frac{80}{x^2}=0$

$8x^3=80$

$x^3=10$

$x=\sqrt[3]{10}\approx 2.15$ m

Then $h=\frac{10}{x^2}\approx 2.15$ m

Step 6: For minimum cost, the container should have a square base approximately 2.15 m × 2.15 m and height approximately 2.15 m (essentially a cube). This minimizes the total material cost while maintaining the required volume. The expensive base is minimized while still providing adequate volume.

Light and Reflection Problems

Example 8:

A light is 5 meters above the ground. A person 2 meters tall walks away from the light at 1.5 m/s. How fast is the tip of the person's shadow moving along the ground when the person is 10 meters from the base of the light?

[Note: This is actually a related rates problem but included to show connection]

Interpretation focus: The answer would represent the speed at which the shadow's endpoint moves, which is faster than the person walks because the shadow lengthens as the person moves away.

Common Pitfalls and Checks

Verifying Your Answer

Always check:

Example 9:

If optimizing a rectangle's dimensions gives negative values, this indicates an error—dimensions must be positive.

Practice Section