Applied Optimization in Business Calculus

Applied Optimization

Introduction to Applied Optimization

Applied optimization involves finding the maximum or minimum values of a function that models a real-world scenario, subject to certain constraints. This process is essential in business calculus for solving problems related to maximizing profit, minimizing cost, or optimizing resource allocation.

• Optimization is the process of making something as effective or functional as possible within given constraints.

• To solve an optimization problem, express the quantity to be optimized as a function of one variable, determine the domain, and use calculus to find critical points and endpoints.

• Critical points occur where the first derivative is zero or undefined.

• The second derivative test or evaluation at endpoints determines whether these points are maxima or minima.

General Steps for Solving Optimization Problems

Step-by-Step Method

Follow these steps to solve applied optimization problems:

1. Draw a diagram and identify all relevant variables.

2. Write the function to be optimized (objective function).

3. Express the function in terms of a single variable using constraints (may require additional equations).

4. Determine the domain of the variable based on physical or contextual restrictions.

5. Find critical points by setting the first derivative equal to zero: $f'(x) = 0$.

6. Test endpoints (if the interval is closed) or use the second derivative test (if the interval is open):

   • If $f''(x) < 0$, the critical point is a maximum.

   • If $f''(x) > 0$, the critical point is a minimum.

Examples of Applied Optimization

1. Maximizing Area with Fencing

Given a fixed amount of fencing, determine the dimensions of a rectangular enclosure (with one side formed by a rock wall) that maximize the area.

• Objective function: Area $A = x \cdot y$

• Constraint: $2x + y = 200$ (if $x$ is the length parallel to the wall and $y$ is perpendicular)

• Express $A$ in terms of one variable: $y = 200 - 2x \implies A(x) = x(200 - 2x)$

• Find maximum by solving: $A'(x) = 0$

2. Minimizing Surface Area of a Box (No Lid)

Find the dimensions of a rectangular box with a square base and no lid that minimize the surface area for a given volume.

• Let $x$ = side of base, $y$ = height

• Volume constraint: $x^2y = 125$

• Surface area: $A = x^2 + 4xy$

• Express $y$ in terms of $x$: $y = \frac{125}{x^2}$

• Surface area as a function of $x$: $A(x) = x^2 + 4x\left(\frac{125}{x^2}\right) = x^2 + \frac{500}{x}$

• Find minimum by solving: $A'(x) = 0$

3. Maximizing Printed Area of a Poster

Given a fixed total area and specific margins, determine the dimensions that maximize the printed area.

• Total area constraint: $xy = 1150$

• Printed area: $A_p = (x-4)(y-5)$ (subtracting margins)

• Express $y$ in terms of $x$: $y = \frac{1150}{x}$

• Printed area as a function of $x$: $A_p(x) = (x-4)\left(\frac{1150}{x} - 5\right)$

• Find maximum by solving: $A_p'(x) = 0$

4. Maximizing Revenue and Profit

To maximize revenue or profit, use the price-demand function and cost function to express revenue and profit in terms of the number of items sold.

• Price-demand function: $p(x) = 100 - \frac{x}{4}$

• Revenue function: $R(x) = p(x) \cdot x = (100 - \frac{x}{4})x = 100x - \frac{x^2}{4}$

• Profit function: $P(x) = R(x) - C(x)$

• Find maximum by solving: $R'(x) = 0$ or $P'(x) = 0$

• Domain restrictions: $x > 0$, $p(x) > 0$

5. Maximizing Profit: Custom Phone Cases Example

• Price function: $p(x) = 50 - 0.2x$

• Cost function: $C(x) = 100 + 15x$

• Revenue: $R(x) = p(x) \cdot x$

• Profit: $P(x) = R(x) - C(x)$

• Find $x$ that maximizes $P(x)$ by solving: $P'(x) = 0$

6. Maximizing Revenue: Latte Pricing Example

• Given a linear relationship between price and customers, express the number of items sold as a function of price.

• Revenue function: $R(p) = p \cdot n(p)$, where $n(p)$ is the number of customers at price $p$.

• Find the price $p$ that maximizes $R(p)$ by solving $R'(p) = 0$.

7. Maximizing Light Through a Norman Window

A Norman window consists of a rectangle topped with a semicircle. Given a fixed perimeter, find the dimensions that maximize the amount of light passing through, considering different light transmission rates for the rectangle and semicircle.

• Let $x$ = width, $y$ = height of rectangle

• Perimeter constraint: $2y + x + \frac{\pi x}{2} = 24$

• Light function: $L(x, y) = xy + \frac{1}{4}\left(\frac{\pi x^2}{8}\right)$ (if semicircle transmits 1/4 as much light)

• Express $y$ in terms of $x$ using the constraint, substitute into $L(x, y)$, and maximize.

8. Maximizing Volume: Packaging Design

Find the size of the squares to cut from the corners of a rectangular sheet to maximize the volume of the resulting open-top box.

• Let $x$ = side of cut squares

• Original sheet: $3$ft by $8$ ft

• Box dimensions: $(8-2x)$ by $(3-2x)$ by $x$

• Volume function: $V(x) = x(8-2x)(3-2x)$

• Find maximum by solving: $V'(x) = 0$

• Domain: $0 < x < 1.5$ (since $3-2x > 0$ and $8-2x > 0$)

9. Shortest Rope Problem

Given a tree and a cliff, find the shortest rope that can be tied from the ground, over the top of the tree, to the top of the cliff.

• Let $x$ = horizontal distance from the base of the tree to the point where the rope touches the ground

• Use the Pythagorean theorem to express the total rope length as a function of $x$

• Minimize the total length by taking the derivative and setting it to zero

Additional info: The above examples illustrate the application of calculus-based optimization to real-world business and design problems, a core topic in business calculus. The images included directly support the geometric visualization of the problems discussed.