Question

Suppose a business can sell x gadgets for p(x)=250 - 0.01x dollars a piece, and it costs the business c(x) = 1,000 + 25x dollars so produce the x gadgets. Determine the production level and cost per gadget required to maximize profit.
A. 13,750 gadget at $112.50 each
B. 111 gadgets at $248.89 each
C. 11, 250 gadgets at $137.50 each
D. 10,000 gadgets at $150.00 each

Answer 1

Answer:

A

Explanation:

In this question, we are asked to determine the production level and cost per gadget required for profit maximization.

p(x) = 250 - 0.01x

c(x) = 1000+25x

The total amount made from selling a number of x gadgets= number of gadgets * cost at which the gadget is sold per piece

Mathematically = r(x) = x p(x) = 250x - 0.01x^2

The profit made is equate to the revenue minus the cost of production

Mathematically

p(x) = r(x) - c(x) = 250x - 0.01x^2 - (1000+25x) = 225x - 0.01x^2 -1000

To get maximum profit, the first derivative of the profit must be equal to 0

p'(x) = 225 - 0.02x

p'(x)=0 gives 225-0.02x=0

=> x = 11250

and p(x) = 250 - 0.01*11250 = 137.50

Answer 2

Answer:

C. 11, 250 gadgets at $137.50 each

Explanation:

The profit function for x units sold is given by the difference between the revenue (price multiplied by number of units) and cost functions.

$P(x) = xp(x) - c(x) = 250x-0.01x^x -1,000 -25x\\P(x) = -0.01x^2+225x-1,000$

The output level 'x' for which the derivate of the profit function is zero maximizes profit:

$P'(x) =0= -0.02x+225\\x=11,250\ gadgets$

The price at a production level of x = 11,250 gadgets is:

$p(11,250) = 250-0.01*11,250\\p(11,250) = \ 137.50$

Therefore, the answer is C. 11, 250 gadgets at $137.50 each