7. 1/2 points | Previous Answers My Notes Ask Your Teache A firm that produces I

Question

7. 1/2 points | Previous Answers My Notes Ask Your Teache A firm that produces Items has monthly average costs, in dollars per Item, given by the following function where q is the number of Items produced per month. AC(q)- 38,000200+q The firm can sell Items in a competitive market for $1800 per Item. If production is limited to 250 Items per month, find the number of Items that gives maximum profit, and find the maximum profit. q=250 Items maximum profit $ 235900.79 X Submit Answer Save Progress 03 points | Previous Arnswers My Notes Ask Your Teache You sell Things. The weekly demand function ror 9 Things is 400-dollars per Thing, and the average cost or production and sale is AC(Q) = 300 + 2q dollars per Thing (a) Find the quantity that will maximize profit. 16.54 Things (b) Find the selling price at this optimal quantity 201 x per Thing (c) What is the maximum profit? 3101.277 X 9. 0V3 points | Prewous Answers My Notes Ask Your Teache A large corporation with monopolistic control in the marketplace has its average daily costs, in dollars per unit, given by the following. 700 : daily demand for q units or its product is given by = 180,000 for its product? x= 100 50q dollars per unit. Find the quantity that gives maximum profit, and find the maximum profit what selling price should the corporation Xunits maximum profit $ selling price

Explanation / Answer

For #7

To find maximum profit, find revenue - cost = profit function

Profit = 1800q - 38000/q -200-q

Now, find derivative of this function.

P'(q)= 1800+38000q^-2-1

Now, plug in 'q' as 250

P'(250)= 1800+38000(250)^-2 -1 =$1799.61

For #8

Revenue = q(400-1/2q)

Distribute

Revenue =400q -0.5q^2

Cost =300+2q

So, profit = Revenue-cost

Profit = 400q-0.5q^2 -300-2q

Simplify

P(q) = -0.5q^2 +398q-300

Now, find derivative of this function.

p'(q)= -1q +398

Now, set -1q+398 =0 and solve for 'q'

Subtract both sides 398

-1q= -398

Divide both sides by -1

q =398

So,answer for #8a is 398.

For #8b

Selling price = cost price +profit.

So, Selling price = 300+2q-0.5q^2 +398q-300

                           = -0.5q^2 +400q

So, selling price at this optimal quantity q = 398 is

=-0.5(398)^2 +400(398)=79998

So, selling price =$79998

For #8C

Maximum profit is plug in the 398 for q in profit function.

P(398)= -0.5(398)^2 +398(398)-300=$78902

So, maxium profit = $78902.

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