Answer: To maximize profit, we need to find the level of output and unit price where the difference between revenue and cost is maximized. Profit (π) is given by the equation:
π=TR−TC
π=TR−TC
where TRTR is total revenue and TCTC is total cost.
Given:
TR=P⋅Q=(10−0.003Q)Q=10Q−0.003Q2TR=P⋅Q=(10−0.003Q)Q=10Q−0.003Q2
TC=1000+3Q+0.004Q2TC=1000+3Q+0.004Q2
We can now find the level of output and the unit price at which the profit is maximized.
a. To find the level of output and the unit price at which profit is maximized, we need to find the derivative of the profit function with respect to quantity (Q), set it equal to zero, and solve for Q.
dπdQ=d(TR−TC)dQ=d(TR)dQ−d(TC)dQ=0dQdπ=dQd(TR−TC)=dQd(TR)−dQd(TC)=0
d(TR)dQ=d(10Q−0.003Q2)dQ=10−0.006QdQd(TR)=dQd(10Q−0.003Q2)=10−0.006Q
d(TC)dQ=d(1000+3Q+0.004Q2)dQ=3+0.008QdQd(TC)=dQd(1000+3Q+0.004Q2)=3+0.008Q
Setting the derivatives equal to each other:
10−0.006Q=3+0.008Q10−0.006Q=3+0.008Q
0.014Q=70.014Q=7
Q=70.014Q=0.0147
Q=500Q=500
Now, to find the unit price (P), we can use the demand function:
P=10−0.003QP=10−0.003Q
P=10−0.003(500)P=10−0.003(500)
P=10−1.5P=10−1.5
P=8.5P=8.5
So, the level of output (Q) at which profit is maximized is 500 units, and the unit price (P) at which profit is maximized is $8.5.
b. To calculate the amount of profit at this level, we substitute the value of Q into either the total revenue or total cost function and then calculate profit using the formula:
π=TR−TCπ=TR−TC
Using the total revenue function:
TR=P⋅Q=8.5⋅500=4250TR=P⋅Q=8.5⋅500=4250
π=TR−TC=4250−(1000+3(500)+0.004(500)2)π=TR−TC=4250−(1000+3(500)+0.004(500)2)
π=4250−(1000+1500+0.004(250000))π=4250−(1000+1500+0.004(250000))
π=4250−(1000+1500+1000)π=4250−(1000+1500+1000)
π=4250−3500π=4250−3500
π=750π=750
So, the amount of profit at this level is $750.
