Business Calculus: Maximizing Revenue with Price-Demand Functions

Q19. A company manufactures and sells x smartphones per week. The weekly price-demand and cost equations are, respectively, $p = 500 - 0.5x$ and $c(x) = 20,000 + 135x$.

• A1) What price should the company charge for phones to maximize weekly revenue?

• A2) How many phones should be produced to maximize the weekly revenue?

• A3) What is the maximum weekly revenue?

Background

Topic: Revenue Maximization using Price-Demand Functions

This question tests your understanding of how to use price-demand equations to find the price and quantity that maximize revenue, a common application of calculus in business.

Key Terms and Formulas

• Price-demand equation: Relates the price ($p$) at which a product can be sold to the quantity ($x$) sold.

• Revenue function: $R(x) = x \cdot p(x)$, where $p(x)$ is the price-demand equation.

• To maximize revenue: Find the value of $x$ that maximizes $R(x)$, usually by finding where $R'(x) = 0$.

Step-by-Step Guidance

1. Write the revenue function in terms of $x$ using the price-demand equation: $R(x) = x \cdot (500 - 0.5x)$.

2. Expand and simplify the revenue function: $R(x) = 500x - 0.5x^2$.

3. To find the maximum, take the derivative of $R(x)$ with respect to $x$ and set it equal to zero: $R'(x) = 0$.

4. Solve the resulting equation for $x$ to find the quantity that maximizes revenue.

5. Once you have $x$, substitute it back into the price-demand equation to find the price $p$ that should be charged.

6. To find the maximum revenue, substitute the maximizing $x$ value into the revenue function $R(x)$.

Try solving on your own before revealing the answer!

Final Answers:

• A1) Price to charge: $p = 375$

• A2) Number of phones to produce: $x = 250$

• A3) Maximum weekly revenue: $R = 93,750$

We set up the revenue function $R(x) = 500x - 0.5x^2$, took its derivative, set it to zero, and solved for $x$. Plugging $x$ back into the price-demand equation gave the optimal price, and substituting $x$ into $R(x)$ gave the maximum revenue.

Q20. A company manufactures and sells x digital cameras per week. The weekly price-demand and cost equations are, respectively, $p = 40 - 0.8x$ and $c(x) = 20 + 16x$.

• A1) What price should the company charge for cameras to maximize weekly revenue?

• A2) How many cameras should be produced to maximize the weekly revenue?

• A3) What is the maximum weekly revenue?

Background

Topic: Revenue Maximization using Price-Demand Functions

This question is similar to the previous one, focusing on maximizing revenue using a linear price-demand equation and applying calculus techniques.

Key Terms and Formulas

• Price-demand equation: $p = 40 - 0.8x$

• Revenue function: $R(x) = x \cdot p(x)$

• To maximize revenue: Find $x$ where $R'(x) = 0$

Step-by-Step Guidance

1. Write the revenue function: $R(x) = x \cdot (40 - 0.8x)$.

2. Expand and simplify: $R(x) = 40x - 0.8x^2$.

3. Take the derivative of $R(x)$ with respect to $x$ and set it equal to zero: $R'(x) = 0$.

4. Solve for $x$ to find the quantity that maximizes revenue.

5. Substitute this $x$ value into the price-demand equation to find the optimal price $p$.

6. Plug the maximizing $x$ into $R(x)$ to find the maximum revenue.

Try solving on your own before revealing the answer!

Final Answers:

• A1) Price to charge: $p = 20$

• A2) Number of cameras to produce: $x = 25$

• A3) Maximum weekly revenue: $R = 500$

We set up the revenue function $R(x) = 40x - 0.8x^2$, took its derivative, set it to zero, and solved for $x$. Plugging $x$ back into the price-demand equation gave the optimal price, and substituting $x$ into $R(x)$ gave the maximum revenue.