Question #244416

BeachWear Inc. manufactures beach sandals. Their expense and revenue functions are E=-300p + 32,000 and R= -275p^2 + 6,500p . Determine the maximum profit

to the nearest cent.

a. $10,036.36 b. $6764.45 c. $6409.09 d. $2945.45


1
Expert's answer
2021-10-04T17:08:53-0400

Given,

Expense function:

E=300p+32,000E=−300p+32,000

Revenue function:

R=275p2+6500pProfit(π)=REπ=275p2+6,500p(300p+32,000)π=275p2+6,500p+300p32,000π=275p2+6,800p32,000R=−275p^2+6500p\\ Profit(π)=R−E\\π=−275p^2+6,500p−(−300p+32,000)\\π=−275p^2+6,500p+300p−32,000\\π=−275p^2+6,800p−32,000

To maximize the profit function let us derivate the function with

 respect to p

dπdp=2×275p+6,800Set,dπdp=02×275p+6,800=06800=2×275pp=6800550p=12.3636\frac{dπ}{dp}=−2×275p+6,800\\Set,\\ \frac{dπ}{dp}=0\\−2×275p+6,800=0\\6800=2×275p\\p=\frac{6800}{550}\\p=12.3636

To check whether the p=12.3636 will give maximum profit or not let us find

 the second−order derivative

d2πdπ2=2×275d2πdπ2=550<0\frac{d^2π}{dπ^2}=−2×275\\\frac{d^2π}{dπ^2}=−550<0

Since the value of second−order derivative is less than zero p=12.3636 will

 give maximum profit

At p=12.3636 the profit will be maximum.

At p=12.3636 the value of revenue will be a sgiven below

R=275×(12.3636)2+6,500×12.3636R=42,036.1164+8036.34R=38,327.2836R=−275×(12.3636)^2+6,500×12.3636\\R=−42,036.1164+8036.34\\R=38,327.2836

At p=12.3636 the value of expense will be a sgiven below

E=300××12.3636+32,000E=3,709.08+32,000E=$28,290.92So,Profit=REProfit=38,327.283628,290.92Profit=$10,036.36E=−300××12.3636+32,000\\E=−3,709.08+32,000\\E=\$28,290.92\\So,\\Profit=R−E\\Profit=38,327.2836−28,290.92\\Profit=\$10,036.36


The correct option is A.

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