Asked by Jacke
1. For the given cost and demand functions, find the production level that will maximize profit.
C(x)=820+5x+0.01x^2, p(x)=18-x/400. Find the answer to the nearest whole number.
a. x = 538
b. x = 525
c. x = 511
d. x = 520
2. A manufacturer has been selling 1,100 television sets a week at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. How large a rebate should the company offer the buyer in order to maximize its revenue?
Select one:
a. $127.50
b. $85.00
c. $170.00
d. $340.00
C(x)=820+5x+0.01x^2, p(x)=18-x/400. Find the answer to the nearest whole number.
a. x = 538
b. x = 525
c. x = 511
d. x = 520
2. A manufacturer has been selling 1,100 television sets a week at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week. How large a rebate should the company offer the buyer in order to maximize its revenue?
Select one:
a. $127.50
b. $85.00
c. $170.00
d. $340.00
Answers
Answered by
Steve
profit is revenue minus cost, or
x*p(x) - c(x)
= x(18-x/400)-(820+5x+0.01x^2)
= -0.0125x^2 + 13x - 820
That's just a parabola, with its vertex at x = -b/2a = 520
work #2 the same way. Figure a quantity and a price in terms of x, the number of $10 rebates. Then, revenue is price*quantity.
x*p(x) - c(x)
= x(18-x/400)-(820+5x+0.01x^2)
= -0.0125x^2 + 13x - 820
That's just a parabola, with its vertex at x = -b/2a = 520
work #2 the same way. Figure a quantity and a price in terms of x, the number of $10 rebates. Then, revenue is price*quantity.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.