Question

WidgCo is a company that produces and sells widgets. Let p denote the price per widget (measured in dollars), and let x be the monthly demand for widgets. WidgCo's marketing department determines that p and x are related by the following demand equation: p(x) = 211 − 5x. The cost of producing x widgets is given by C(x) = 1,401 + 16x. Construct the revenue function, R(x). Find the production level that will result in the maximum revenue. Find the maximum monthly revenue. Enter your answer to part c in the box below. Round your answer to the nearest cent.

Answer 1

Answer:

The answer to this question can be defined as follows:

i) $\bold{p(x)= -5x^2+ 217x- 1,401}$

ii) x=22 units

iii) Maximum profit: 953.45

Step-by-step explanation:

Given value:

p(x) = 211 − 5x \\

C(x) = 1,401 + 16x

The formula for calculating the profit function value:

$\to \bold{P(x)= xp(x) - C(x)}$

             $= x(211-5x)- 1,401 + 16x\\\\= 211x -5x^2-1,401+16x\\\\= -5x^2+ 217x- 1,401$

The formula for calculating the value for maximum profit:

$p'(x) = - 10x + 217 =0$

$217= 10x\\\\x= \frac{217}{10}\\\\x= 21.7$

So, the production level is 22 units

Maximum profit:

$\to p(21.7)= - 5 (21.7)^2+217 (21.7) -1,401$

           $=-5 (470.89) +217 (21.7) -1,401\\\\=-2354.45+ 4708.9-1,401\\\\= 953.45$

Answer 2

The maximum monthly revenue of WidgCo is; $489

• We are given;

price function; p(x) = 211 - 5x

Cost function; C(x) = 1401 + 16x

• We know that revenue would be gotten by the formula;

Revenue = number of units sold × price per unit

Thus;

R(x) = x(211 - 5x)

Expanding this gives;

R(x) = 211x - 5x²

• Maximum revenue will be the revenue for the number of widgets when R'(x) = 0. Thus;

R'(x) = 211 - 10x

At R'(x) = 0, we have;

211 - 10x = 0

x = 211/10

x = 21.1

x ≈ 21 widgets

• Thus;

Maximum Revenue is;

R(20) = 195(21) - 1401 - 5(21)²

R(20) = $489

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