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# MTH132 - Section 4.4 - Optimization ## What is Optimization? **Optimization** problems involve finding the **absolute maximum** or **absolute minimum** value of a function on a given interval. ## Terminology - **Objective Function:** The function we want to maximize or minimize. - **Constraint:** Side conditions or equations that limit the possible values of the variables. ## Strategy for Solving Optimization Problems 1. **Read the problem carefully.** Make sure you understand what is being asked. 2. **Draw a diagram** if possible. 3. **Introduce notation.** Assign symbols to all quantities that are functions of time. 4. **Express the objective function** in terms of the symbols. 5. **Use the given information** (constraints) to express the objective function in terms of *one* variable. 6. **Find the absolute maximum or minimum** value of the objective function on the interval determined by the constraints. - Find the critical points of the function. - Evaluate the function at the critical points and endpoints. - The largest value is the absolute maximum, and the smallest value is the absolute minimum. 7. **Answer the question.** Be sure to include units. ## Example 1 A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area? ### Solution 1. Read the problem. 2. Draw a diagram: ``` __________ | | | | w | A | | | |__________| l River ``` 3. Introduce notation: - $A =$ Area of the field - $l =$ Length of the field - $w =$ Width of the field 4. Express the objective function in terms of the symbols: - Maximize: $A = lw$ 5. Use the given information (constraints) to express the objective function in terms of *one* variable: - Constraint: $2w + l = 2400 \implies l = 2400 - 2w$ - Substitute into the objective function: $A = (2400 - 2w)w = 2400w - 2w^2$ 6. Find the absolute maximum value of the objective function on the interval determined by the constraints: - Domain: $0 \le w \le 1200$ - $A'(w) = 2400 - 4w$ - $A'(w) = 0 \implies 2400 - 4w = 0 \implies w = 600$ - Critical point: $w = 600$ - Evaluate $A$ at the critical point and endpoints: - $A(0) = 0$ - $A(600) = 2400(600) - 2(600)^2 = 720000$ - $A(1200) = 0$ 7. Answer the question: - The maximum area is $720,000 \text{ ft}^2$ when $w = 600 \text{ ft}$ and $l = 2400 - 2(600) = 1200 \text{ ft}$. ## Example 2 Find the point on the parabola $y^2 = 2x$ that is closest to the point $(1, 4)$. ### Solution 1. Read the problem. 2. Draw a diagram: ``` (1, 4). | | | * (x, y) /|\ | | ------------ parabola ``` 3. Introduce notation: - $(x, y) =$ Point on the parabola - $d =$ Distance between $(x, y)$ and $(1, 4)$ 4. Express the objective function in terms of the symbols: - Minimize: $d = \sqrt{(x - 1)^2 + (y - 4)^2}$ 5. Use the given information (constraints) to express the objective function in terms of *one* variable: - Constraint: $y^2 = 2x \implies x = \frac{y^2}{2}$ - Substitute into the objective function: $d = \sqrt{(\frac{y^2}{2} - 1)^2 + (y - 4)^2}$ - To simplify, minimize $d^2$ instead of $d$: $d^2 = (\frac{y^2}{2} - 1)^2 + (y - 4)^2$ 6. Find the absolute minimum value of the objective function on the interval determined by the constraints: - $f(y) = (\frac{y^2}{2} - 1)^2 + (y - 4)^2$ - $f'(y) = 2(\frac{y^2}{2} - 1)(y) + 2(y - 4) = y^3 - 2y + 2y - 8 = y^3 - 8$ - $f'(y) = 0 \implies y^3 - 8 = 0 \implies y^3 = 8 \implies y = 2$ - Critical point: $y = 2$ - Since $f'(y) < 0$ for $y < 2$ and $f'(y) > 0$ for $y > 2$, $y = 2$ is a local minimum. - Since there is only one critical point, it must be the absolute minimum. 7. Answer the question: - $x = \frac{y^2}{2} = \frac{2^2}{2} = 2$ - The point on the parabola closest to $(1, 4)$ is $(2, 2)$.

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