Ule 2: HW 7 Section 2.7 YES CRADED
The profit (in dollars) from the sale of lawn mowers is .
(A) Find the average profit per mower if 50 mowers are produced.
(B) Find the marginal average profit at a production level of 50 mowers, and interpret the results.
© Use the results from parts (A) and (B) to estimate the average profit per mower if 51 mowers are produced.
(A) If 50 mowers are produced, the average profit per mower is .
(Round to the nearest cent as needed.)
(B) The marginal average profit at a production level of 50 units is per mower.
(Round to the nearest cent as needed.).

Ask by Paul Norton. in the United States

Mar 30,2025

Question

Ule 2: HW 7 Section 2.7 YES CRADED
The profit (in dollars) from the sale of lawn mowers is .
(A) Find the average profit per mower if 50 mowers are produced.
(B) Find the marginal average profit at a production level of 50 mowers, and interpret the results.
© Use the results from parts (A) and (B) to estimate the average profit per mower if 51 mowers are produced.
(A) If 50 mowers are produced, the average profit per mower is .
(Round to the nearest cent as needed.)
(B) The marginal average profit at a production level of 50 units is per mower.
(Round to the nearest cent as needed.).

Ask by Paul Norton. in the United States

Mar 30,2025

UpStudy AI · Accepted Answer

**(A)**  
The profit function is given by  
$$
P(x) = 70x - 0.07x^2 - 450.
$$  
For 50 mowers, the total profit is  
$$
P(50) = 70(50) - 0.07(50)^2 - 450.
$$  
Calculate each term:  
$$
70(50) = 3500,
$$  
$$
(50)^2 = 2500 \quad \text{so} \quad 0.07(2500) = 175,
$$  
thus  
$$
P(50) = 3500 - 175 - 450 = 2875.
$$  
The average profit per mower is  
$$
\text{Average profit} = \frac{P(50)}{50} = \frac{2875}{50} = 57.50 \text{ dollars}.
$$

**(B)**  
Let the average profit per mower be  
$$
A(x) = \frac{P(x)}{x} = \frac{70x - 0.07x^2 - 450}{x}.
$$  
Simplify $$ A(x) $$:  
$$
A(x) = 70 - 0.07x - \frac{450}{x}.
$$  
The marginal average profit is the derivative $$ A'(x) $$. Differentiate term by term:

- The derivative of $$ 70 $$ is $$ 0 $$.  
- The derivative of $$ -0.07x $$ is $$ -0.07 $$.  
- The derivative of $$ -\frac{450}{x} = -450x^{-1} $$ is  
  $$
  -450(-1)x^{-2} = \frac{450}{x^2}.
  $$
Thus,  
$$
A'(x) = -0.07 + \frac{450}{x^2}.
$$  
Evaluate at $$ x = 50 $$:  
$$
A'(50) = -0.07 + \frac{450}{50^2} = -0.07 + \frac{450}{2500}.
$$  
Compute the fraction:  
$$
\frac{450}{2500} = 0.18.
$$  
So,  
$$
A'(50) = -0.07 + 0.18 = 0.11.
$$  
This means that at a production level of 50 mowers, producing one additional mower (i.e., moving from 50 to 51 mowers) will increase the average profit per mower by approximately $$ \$0.11 $$.

**(C)**  
Using the marginal average profit as a linear approximation, the average profit per mower for 51 mowers is estimated by  
$$
A(51) \approx A(50) + A'(50) \times 1.
$$  
We already have  
$$
A(50) = 57.50 \quad \text{and} \quad A'(50) = 0.11.
$$  
Thus,  
$$
A(51) \approx 57.50 + 0.11 = 57.61 \text{ dollars per mower}.
$$
**(A)** If 50 mowers are produced, the average profit per mower is **\$57.50**. **(B)** The marginal average profit at a production level of 50 mowers is **\$0.11** per mower. **(C)** Estimating the average profit per mower for 51 mowers is **\$57.61**.

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