SOLUTION:
5.​A company that builds phones earns a profit on sales according to the quadratic relation
P = -10n2 + 1000n -2500
where P is the profit
Algebra.Com
Question 720485:
5.A company that builds phones earns a profit on sales according to the quadratic relation
P = -10n2 + 1000n -2500
where P is the profit earned in dollars, and N is the number of crates of phones sold.
a. How many crates must be sold to make a maximum profit?
b. What is the maximum profit that can be made?
c. If you sell 11 crates how will you make a profit or a loss? How much?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
5.A company that builds phones earns a profit on sales according to the quadratic relation
P = -10n2 + 1000n -2500
where P is the profit earned in dollars, and N is the number of crates of phones sold.
a. How many crates must be sold to make a maximum profit?
P = -10n^2 + 1000n -2500
Find derivative:
P' = -20n + 1000
Set P' to zero to find critical points:
0 = -20n + 1000
-1000 = -20n
1000 = 20n
100 = 2n
50 = n
.
b. What is the maximum profit that can be made?
plug n=50 and find P:
P = -10n^2 + 1000n -2500
P = -10(50)^2 + 1000(50) -2500
P = -10(2500) + 1000(50) -2500
P = -25000 + 50000 -2500
P = 25000 -2500
P = $22500
.
c. If you sell 11 crates how will you make a profit or a loss? How much?
plug n=11 and find P:
P = -10n^2 + 1000n -2500
P = -10(11)^2 + 1000(11) -2500
P = -10(121) + 11000 -2500
P = -1210 + 11000 -2500
P = 9790 -2500
P = $7290 (Profit)
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