Applied Optimization: Videos & Practice Problems

To solve applied optimization problems, express the value to be optimized as a function, identify constraints, and determine the domain. For example, maximizing area with a fixed perimeter involves finding critical points through the first derivative and applying the extreme value theorem for closed intervals. In revenue maximization, revenue is calculated as price times quantity sold, with domain restrictions ensuring non-negative values. Understanding these concepts is crucial for effectively addressing real-world scenarios in calculus.

Intro to Applied Optimization: Maximizing Area

In applied optimization problems, we often seek to maximize or minimize a specific value, such as area, under given constraints. A common scenario involves using a fixed amount of resources, like fencing, to create a structure with the largest possible area. For instance, consider a problem where we have 200 feet of fencing to construct a rectangular area, with one side against a rock wall, meaning we only need to fence three sides.

To solve this, we start by defining the dimensions of the rectangle. Let’s denote the length as \( x \) and the width as \( y \). The area \( A \) of the rectangle can be expressed as:

\[ A = x \cdot y \]

Given the constraint of 200 feet of fencing, we can express the perimeter of the three sides as:

\[ 2x + y = 200 \]

From this equation, we can isolate \( y \):

\[ y = 200 - 2x \]

Substituting this expression for \( y \) back into the area function gives us:

\[ A = x(200 - 2x) = 200x - 2x^2 \]

Next, we need to determine the domain of our function. Since dimensions cannot be negative, we have:

\[ 0 \leq x \leq 100 \]

To find the maximum area, we calculate the first derivative of the area function:

\[ A' = 200 - 4x \]

Setting the derivative equal to zero to find critical points:

\[ 200 - 4x = 0 \implies x = 50 \]

Now, we evaluate the area at the critical point and the endpoints of the interval (0 and 100) to apply the Extreme Value Theorem:

\[ A(0) = 0 \]

\[ A(100) = 0 \]

\[ A(50) = 200(50) - 2(50^2) = 5000 \]

Thus, the maximum area is 5000 square feet, occurring when \( x = 50 \) feet. To find the corresponding width \( y \), we substitute \( x \) back into the equation for \( y \):

\[ y = 200 - 2(50) = 100 \]

Therefore, the dimensions that yield the maximum area are 50 feet by 100 feet.

In summary, when tackling applied optimization problems, it is essential to define the variables, express the function to be optimized, determine the domain, find critical points, and evaluate the function at these points and the endpoints to identify maximum or minimum values. If the domain is closed, use the Extreme Value Theorem; if open, apply the second derivative test to ascertain concavity and determine maxima or minima.

Example 1: Minimizing Surface Area

In applied optimization problems, one common task is to minimize the surface area of a three-dimensional object. For instance, consider a shipping company that needs to construct a rectangular box with a square base, designed to hold a volume of 125 cubic inches. The goal is to determine the dimensions that will minimize the amount of cardboard used, which directly correlates to minimizing the surface area of the box.

To begin, we define the variables: let x represent the length of each side of the square base, and y denote the height of the box. The surface area S of the box, which has no lid, can be expressed as:

$$ S = 4xy + x^2 $$

Here, 4xy accounts for the four sides of the box, and x² represents the area of the base. Given the constraint of a fixed volume, we know:

$$ V = x^2y = 125 $$

From this equation, we can solve for y in terms of x:

$$ y = \frac{125}{x^2} $$

Substituting this expression for y back into the surface area equation gives us:

$$ S = 4x\left(\frac{125}{x^2}\right) + x^2 = \frac{500}{x} + x^2 $$

Next, we need to find the critical points of this function to determine where the surface area is minimized. To do this, we calculate the derivative of S:

$$ S' = -\frac{500}{x^2} + 2x $$

Setting the derivative equal to zero to find critical points:

$$ -\frac{500}{x^2} + 2x = 0 $$

Rearranging gives:

$$ \frac{500}{x^2} = 2x $$

Multiplying both sides by x² leads to:

$$ 500 = 2x^3 $$

Dividing by 2 results in:

$$ x^3 = 250 $$

Taking the cube root yields:

$$ x = \sqrt[3]{250} \approx 6.3 $$

With x determined, we can find y using the earlier equation:

$$ y = \frac{125}{(6.3)^2} \approx 3.15 $$

Thus, the dimensions of the box that minimize the surface area are approximately 6.3 inches for each side of the square base and 3.15 inches for the height. This solution illustrates the process of applying calculus to solve real-world optimization problems effectively.

A poster is set to have a total area of 1150 cm², with 2-cm margins on the sides and the top, and a 3-cm margin at the bottom. What dimensions will maximize the printed area?

• A. Width: 22.5 cm, Height: 51.2 cm
• B. Width: 35.6 cm, Height: 32.3 cm
• C. Width = 30.3 cm, Height = 37.9 cm
• D. Width: 28.1 cm, Height: 41.0 cm

Answer: (C)

Maximizing Profit & Revenue

In applied optimization problems, particularly in business contexts, we often aim to maximize revenue or profit by expressing these values as functions and identifying their maximum or minimum points. To illustrate this, consider a coffee shop that sells x coffees per day at a price of p dollars each. The price demand function is given by:

P(x) = 100 - \frac{x}{4}

To maximize daily revenue, we first need to express revenue as a function of x. Revenue (R) is calculated as the product of the price per coffee and the number of coffees sold:

R(x) = P(x) \cdot x = \left(100 - \frac{x}{4}\right) \cdot x

Distributing x gives us:

R(x) = 100x - \frac{x^2}{4}

This function is now expressed solely in terms of x. Next, we need to consider any domain restrictions. Since x represents the number of coffees sold, it must be non-negative:

x \geq 0

Additionally, the price function must also remain non-negative, leading to the inequality:

100 - \frac{x}{4} \geq 0

Solving this inequality results in:

x \leq 400

Thus, the domain for x is:

0 \leq x \leq 400

With the revenue function established and the domain restrictions identified, the next steps involve finding critical points and applying methods such as the extreme value theorem or the second derivative test to determine the maximum revenue. This structured approach allows for effective problem-solving in maximizing business revenue.

Example 2: Maximizing Revenue

In this analysis of revenue maximization, we begin with the revenue function defined as the product of price and quantity sold, expressed mathematically as:

$$ R(x) = 100x - \frac{x^2}{4} $$

Here, \( x \) represents the number of items sold, with the domain restrictions set to \( 0 \leq x \leq 400 \). This ensures that both the quantity sold and the price remain non-negative.

To find the critical points, we first compute the first derivative of the revenue function:

$$ R'(x) = 100 - \frac{x}{2} $$

Setting the first derivative equal to zero allows us to identify critical points:

$$ 100 - \frac{x}{2} = 0 $$

Solving for \( x \) gives:

$$ x = 200 $$

Next, we evaluate the revenue at the endpoints of the domain and the critical point to determine the maximum revenue. We calculate:

1. \( R(0) = 100(0) - \frac{0^2}{4} = 0 \)

2. \( R(200) = 100(200) - \frac{200^2}{4} = 10,000 \)

3. \( R(400) = 100(400) - \frac{400^2}{4} = 0 \)

From these calculations, we observe that the maximum revenue occurs at the critical point \( x = 200 \), yielding a maximum revenue of:

$$ R(200) = 10,000 $$

This indicates that to achieve the maximum revenue of $10,000, 200 items must be sold.

Alternatively, one could apply the second derivative test to confirm the nature of the critical point. The second derivative is calculated as:

$$ R''(x) = -\frac{1}{2} $$

Since the second derivative is negative, it confirms that the critical point at \( x = 200 \) is indeed a maximum. This method provides an efficient way to verify the results obtained from evaluating the endpoints and critical points.

Example 3: Maximizing Profit

In this scenario, we explore how to maximize profit for a business selling custom-made phone cases. The price per case, denoted as P(x), is influenced by the number of cases sold, represented by the function P(x) = 50 - 0.2x, where x is the number of cases sold and P(x) is the price in dollars. The total cost to produce x cases is given by the cost function C(x) = 100 + 15x.

To find the maximum profit, we first need to establish the profit function, which is defined as the revenue minus the cost. The revenue R can be calculated as the price per case multiplied by the number of cases sold: R(x) = P(x) * x = (50 - 0.2x) * x. Thus, the profit function can be expressed as:

Profit (P) = Revenue - Cost

P(x) = R(x) - C(x) = (50 - 0.2x)x - (100 + 15x)

After simplifying, we arrive at the profit function:

P(x) = -0.2x² + 35x - 100

Next, we determine the domain restrictions. Since selling a negative number of cases is not feasible, we establish that x ≥ 0. Additionally, the price must remain non-negative, leading to the inequality P(x) ≥ 0, which simplifies to x ≤ 250. Therefore, the domain for x is 0 ≤ x ≤ 250.

To find the critical points, we calculate the first derivative of the profit function:

P'(x) = -0.4x + 35

Setting the first derivative equal to zero allows us to find the critical points:

-0.4x + 35 = 0

x = 87.5

Since we cannot sell half a phone case, we round this to the nearest whole number, resulting in x = 88.

To confirm that this point yields a maximum profit, we compute the second derivative:

P''(x) = -0.4

Since the second derivative is negative, it indicates that the profit function is concave down at this point, confirming a maximum. Now, we can calculate the maximum profit by substituting x = 88 back into the profit function:

P(88) = -0.2(88)² + 35(88) - 100 = $1,431.20

Finally, to find the price that corresponds to this maximum profit, we substitute x = 88 into the price function:

P(88) = 50 - 0.2(88) = $32.40

In summary, to maximize profit, Gerald and Katya should sell 88 phone cases, resulting in a maximum profit of $1,431.20, with a selling price of $32.40 per case.

Your café sells lattes for $4 each to 100 customers per day. For every $1 increase in price, you would lose 20 customers. Find the price that maximizes revenue. Hint: The # of items sold is based on the number of customers.

• A. $4.00
• B. $4.50
• C. $5.00
• D. $5.50

Answer: (B)

Example 4: Norman Window

The Norman window is a unique shape consisting of a rectangle topped with a semicircle. In optimization problems involving this window, the goal is often to maximize the amount of light that passes through, which can be approached by maximizing the area of the window. Given that the perimeter of the window is 24 meters and the semicircular part allows only one-quarter of the light per square meter compared to the rectangular part, we can derive an area function that accounts for these factors.

To begin, we define the dimensions of the window: let the width of the rectangle be \( x \) and the height be \( y \). The radius of the semicircle is \( \frac{x}{2} \). The total area \( A \) of the window can be expressed as the sum of the area of the rectangle and one-quarter of the area of the semicircle. The area of the rectangle is given by \( A_{rectangle} = x \cdot y \), while the area of the semicircle is \( A_{semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{x}{2}\right)^2 = \frac{\pi x^2}{8} \). Therefore, the area function becomes:

A = xy + \frac{1}{4} \cdot \frac{\pi x^2}{8} = xy + \frac{\pi x^2}{32}.

Next, we need to express \( y \) in terms of \( x \) using the perimeter constraint. The perimeter \( P \) of the window can be expressed as:

P = x + 2y + \frac{\pi x}{2} = 24.

Rearranging this equation to solve for \( y \) gives:

2y = 24 - x - \frac{\pi x}{2},

which simplifies to:

y = 12 - \frac{x}{2} - \frac{\pi x}{4}.

Substituting this expression for \( y \) back into the area function results in:

A = x \left(12 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{32}.

After distributing and simplifying, we obtain a polynomial function in terms of \( x \). To find the critical points, we take the derivative of the area function and set it equal to zero. This will help us identify the values of \( x \) that maximize the area:

A' = 12 - x - \frac{\pi x}{2} + \frac{\pi x}{16} = 0.

Solving this equation yields the critical points. We also need to consider the endpoints of the interval defined by the constraints of the problem, specifically \( 0 \leq x \leq 9.34 \). Evaluating the area function at these critical points and endpoints allows us to determine the maximum area.

After performing these calculations, we find that the optimal width \( x \) is approximately 5.05 meters, leading to a corresponding height \( y \) of about 5.51 meters. Thus, the dimensions that maximize the light passing through the Norman window are 5.05 meters for the width and 5.51 meters for the height.

Example 5: Packaging Design

In applied optimization problems, particularly in packaging design, one common task is to maximize the volume of a box. For instance, consider a box with an open top created by cutting equal-sized squares from the corners of a 3-foot by 8-foot rectangular sheet of cardboard and folding up the sides. The objective is to determine the dimensions of the cut squares that yield the maximum possible volume.

To begin, it's essential to visualize the problem by drawing a diagram and identifying the variables involved. In this case, the side length of the squares cut from each corner is denoted as \( x \). The dimensions of the box after cutting and folding will be influenced by these squares. The length of the box becomes \( 8 - 2x \) and the width becomes \( 3 - 2x \), while the height is simply \( x \). Thus, the volume \( V \) of the box can be expressed as:

\[ V = (8 - 2x)(3 - 2x)(x) \]

Expanding this expression, we find:

\[ V = x(24 - 22x + 4x^2) = 4x^3 - 22x^2 + 24x \]

Next, we need to establish the domain for \( x \). Since \( x \) represents a physical dimension, it must be greater than or equal to 0. Additionally, given the dimensions of the cardboard, \( x \) must also be less than or equal to 1.5 feet, as cutting out squares larger than this would exceed the available material.

To find the critical points for maximizing the volume, we take the derivative of the volume function:

\[ V' = 12x^2 - 44x + 24 \]

Factoring out a 4 gives:

\[ 4(3x^2 - 11x + 6) = 0 \]

Using the quadratic formula, where \( a = 3 \), \( b = -11 \), and \( c = 6 \), we find:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{11 \pm 7}{6} \]

This results in two potential critical points: \( x = 3 \) and \( x = \frac{2}{3} \). However, only \( x = \frac{2}{3} \) (approximately 0.67 feet) is within the valid domain of \( [0, 1.5] \).

To determine whether this critical point yields a maximum volume, we apply the Extreme Value Theorem by evaluating the volume at the endpoints of the interval (0 and 1.5) and at the critical point \( \frac{2}{3} \). Calculating these values:

At \( x = 0 \): \( V(0) = 0 \)

At \( x = 1.5 \): \( V(1.5) = 0 \)

At \( x = \frac{2}{3} \): \( V\left(\frac{2}{3}\right) \approx 7.41 \text{ cubic feet} \)

Since the maximum volume occurs at \( x = \frac{2}{3} \), we conclude that the maximum volume of the box is approximately 7.41 cubic feet, with the dimensions of the squares cut from the corners being \( \frac{2}{3} \) feet.

Example 6: Shortest Rope

In this optimization problem, we aim to determine the shortest length of rope that can be tied from the ground, over the top of a tree, to a cliff. We have a 10-meter tall tree positioned 5 meters away from the base of a rock cliff. To solve this, we will define our variables and derive a function that represents the length of the rope.

First, we denote the length of the rope as \( l \), the height of the cliff as \( h \), and the horizontal distance from the tree to the cliff as \( x \). The relationship between these variables can be expressed using the Pythagorean theorem, where the length of the rope is the hypotenuse of a right triangle formed by the height of the cliff and the horizontal distance. Thus, we have:

\[ l^2 = h^2 + (x + 5)^2 \]

To express \( l \) in terms of a single variable, we need to find \( h \) in terms of \( x \). By analyzing the similar triangles formed by the tree and the cliff, we can set up the ratio:

\[ \frac{h}{10} = \frac{x + 5}{x} \]

Multiplying both sides by \( 10x \) gives us:

\[ h = \frac{10(x + 5)}{x} = 10 + \frac{50}{x} \]

Substituting this expression for \( h \) back into our equation for \( l \), we get:

\[ l = \sqrt{\left(10 + \frac{50}{x}\right)^2 + (x + 5)^2} \]

Next, we need to find the critical points of this function to determine the minimum length of the rope. We start by differentiating \( l \) with respect to \( x \) and setting the derivative equal to zero. Using the chain rule, we find:

\[ l' = \frac{1}{2\sqrt{\left(10 + \frac{50}{x}\right)^2 + (x + 5)^2}} \cdot \left(2\left(10 + \frac{50}{x}\right) \cdot \left(-\frac{50}{x^2}\right) + 2(x + 5)\right) \]

Setting the numerator equal to zero allows us to simplify and solve for \( x \). After some algebraic manipulation, we arrive at:

\[ x^3 + 5x^2 - 500x - 25100 = 0 \]

Factoring this polynomial leads us to the critical points, where we find \( x = \sqrt[3]{500} \) (approximately 7.98) as the only valid solution, since \( x \) must be greater than zero.

To find the minimum length of the rope, we substitute \( x = \sqrt[3]{500} \) back into our equation for \( l \):

\[ l = \sqrt{\left(10 + \frac{50}{\sqrt[3]{500}}\right)^2 + \left(\sqrt[3]{500} + 5\right)^2} \]

Calculating this expression yields a minimum length of approximately 20.81 meters. Therefore, the shortest length of rope that can be tied from the ground over the top of the tree to the cliff is:

20.81 meters

Here’s what students ask on this topic:

How do you solve an applied optimization problem in calculus?

To solve an applied optimization problem, follow these steps: 1) Draw a diagram and identify variables. 2) Write a function for the value being optimized, ensuring it is in terms of one variable. 3) Determine domain restrictions based on the problem's constraints. 4) Find critical points by setting the first derivative equal to zero. 5) Use the extreme value theorem for closed intervals or the second derivative test for open intervals to determine the maximum or minimum value. This process helps in effectively addressing real-world scenarios in calculus.

How do you maximize the area of a rectangular fence with a fixed perimeter?

To maximize the area of a rectangular fence with a fixed perimeter, express the area as a function of one variable. For example, if the perimeter is 200 feet and one side is formed by a rock wall, let the length be x and the width be y. The perimeter constraint is 2x + y = 200. Solve for y: y = 200 - 2x. The area function is A = x(200 - 2x). Find the critical points by setting the first derivative equal to zero: A' = 200 - 4x = 0, giving x = 50. The maximum area is achieved when x = 50 and y = 100, resulting in an area of 5000 square feet.

How do you maximize revenue for a business using calculus?

To maximize revenue, express revenue as a function of the number of items sold. For example, if a coffee shop sells x coffees at p dollars per coffee, and the price demand function is p(x) = 100 - x/4, the revenue function is R(x) = x * p(x) = x(100 - x/4). Simplify to R(x) = 100x - x²/4. Determine domain restrictions: x ≥ 0 and 100 - x/4 ≥ 0, giving x ≤ 400. Find the critical points by setting the first derivative equal to zero: R' = 100 - x/2 = 0, giving x = 200. The maximum revenue is achieved when x = 200.

What is the extreme value theorem and how is it used in applied optimization?

The extreme value theorem states that if a function is continuous on a closed interval [a, b], it has both a maximum and minimum value on that interval. In applied optimization, this theorem is used to find the maximum or minimum values by evaluating the function at critical points and endpoints. For example, if optimizing area with domain restrictions 0 ≤ x ≤ 100, evaluate the area function at x = 0, x = 100, and any critical points within the interval. The largest value is the maximum, and the smallest is the minimum.

How do you determine domain restrictions in applied optimization problems?

To determine domain restrictions in applied optimization problems, consider the constraints given in the problem. For example, if optimizing the area of a fence with a fixed perimeter, the dimensions must be non-negative. If the perimeter is 200 feet, and the length is x, then 2x + y = 200. Since dimensions cannot be negative, x must be between 0 and 100. Similarly, for revenue maximization, ensure the number of items sold and the price are non-negative. These constraints help define the valid range for the variables involved.