Given,
Expense function:
E=−300p+32,000
Revenue function:
R=−275p2+6500pProfit(π)=R−Eπ=−275p2+6,500p−(−300p+32,000)π=−275p2+6,500p+300p−32,000π=−275p2+6,800p−32,000
To maximize the profit function let us derivate the function with
respect to p
dpdπ=−2×275p+6,800Set,dpdπ=0−2×275p+6,800=06800=2×275pp=5506800p=12.3636
To check whether the p=12.3636 will give maximum profit or not let us find
the second−order derivative
dπ2d2π=−2×275dπ2d2π=−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=38,327.2836
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=R−EProfit=38,327.2836−28,290.92Profit=$10,036.36
The correct option is A.
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