Calculus I: Optimization and Related Rates Study Guidance

Q1. Find the two positive numbers with a product of 200 such that the sum of the two numbers is minimized.

Background

Topic: Optimization (Calculus)

This problem asks you to find two positive numbers whose product is fixed (200), and whose sum is as small as possible. This is a classic optimization problem involving constraints.

Key Terms and Formulas

• Objective Function: $S = x + y$ (the sum to minimize)

• Constraint: $xy = 200$

Step-by-Step Guidance

1. Let $x$ and $y$ be the two positive numbers. Write the constraint: $xy = 200$.

2. Express one variable in terms of the other using the constraint. For example, $y = \frac{200}{x}$.

3. Substitute this expression for $y$ into the objective function: $S = x + \frac{200}{x}$.

4. To find the minimum, take the derivative of $S$ with respect to $x$ and set it equal to zero: $\frac{dS}{dx} = 1 - \frac{200}{x^2}$.

Try solving on your own before revealing the answer!

Q2. Find the two positive numbers with a product of 200 such that the sum of the first number and half of the second number is minimized.

Background

Topic: Optimization with Constraints

This problem is similar to Q1, but the objective function is different: you want to minimize $S = x + \frac{1}{2}y$ under the same product constraint.

Key Terms and Formulas

• Objective Function: $S = x + \frac{1}{2}y$

• Constraint: $xy = 200$

Step-by-Step Guidance

1. Let $x$ and $y$ be the two positive numbers. The constraint is $xy = 200$.

2. Express $y$ in terms of $x$: $y = \frac{200}{x}$.

3. Substitute into the objective function: $S = x + \frac{1}{2}\left(\frac{200}{x}\right)$.

4. Simplify: $S = x + \frac{100}{x}$.

5. Take the derivative with respect to $x$ and set it to zero: $\frac{dS}{dx} = 1 - \frac{100}{x^2}$.

Try solving on your own before revealing the answer!

Q3. A farmer needs to build a rectangular pen divided into two rectangular sections. The outer wall costs $5 per foot to build, and the inner dividing wall costs $3 per foot. If the pen must have a total area of 2600 sq ft, find the minimum cost required to build the pen and the dimensions that will achieve the minimum cost.

Background

Topic: Optimization with Cost Functions

This problem involves minimizing the cost of building a pen with a fixed area, where different walls have different costs. You must set up the cost function and use calculus to find the minimum.

Key Terms and Formulas

• Let $x$ and $y$ be the dimensions of the pen.

• Constraint: $xy = 2600$

• Cost Function: $C = 13x + 10y$ (from the problem's solution section)

Step-by-Step Guidance

1. Let $x$ be the length and $y$ the width of the pen. The area constraint is $xy = 2600$.

2. Express $y$ in terms of $x$: $y = \frac{2600}{x}$.

3. Substitute into the cost function: $C = 13x + 10\left(\frac{2600}{x}\right)$.

4. Simplify: $C = 13x + \frac{26000}{x}$.

5. Take the derivative with respect to $x$ and set it to zero: $\frac{dC}{dx} = 13 - \frac{26000}{x^2}$.

Try solving on your own before revealing the answer!

Q4. An open-top box is made by cutting a square from each corner of an 18”×18” sheet of metal and folding up the resulting flaps. What size square should be removed to produce the maximal volume in the box?

Background

Topic: Optimization (Maximizing Volume)

This is a classic box optimization problem. You are asked to maximize the volume of the box formed by cutting squares of side $x$ from each corner and folding up the sides.

Key Terms and Formulas

• Let $x$ be the side length of the square cut from each corner.

• Volume Function: $V = x(18 - 2x)^2$

Step-by-Step Guidance

1. Let $x$ be the side length of the square cut from each corner.

2. After cutting and folding, the box will have height $x$ and base dimensions $(18 - 2x) \times (18 - 2x)$.

3. Write the volume function: $V = x(18 - 2x)^2$.

4. Expand the function if needed: $V = x(324 - 72x + 4x^2)$.

5. Take the derivative with respect to $x$ and set it to zero: $\frac{dV}{dx} = (18 - 2x)^2 + x \cdot 2(18 - 2x)(-2)$.

Try solving on your own before revealing the answer!