SOLUTION: Maximizing profits: In business, profit is the difference between revenue and cost; that is Total Profit=Total Revenue - Total Cost, P(x)=R(x)-C(x), where x is the number of units

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Maximizing profits: In business, profit is the difference between revenue and cost; that is Total Profit=Total Revenue - Total Cost, P(x)=R(x)-C(x), where x is the number of units       Log On


   



Question 945716: Maximizing profits: In business, profit is the difference between revenue and cost; that is Total Profit=Total Revenue - Total Cost, P(x)=R(x)-C(x), where x is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for
R(x)=20x-0.1x^2, C(x)=4x+2

Answer by lwsshak3(11628) About Me  (Show Source):
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Maximizing profits: In business, profit is the difference between revenue and cost; that is Total Profit=Total Revenue - Total Cost, P(x)=R(x)-C(x), where x is the number of units sold. Find the maximum profit and the number of units that must be sold in order to yield the maximum profit for
P(x)=20x-0.1x^2, C(x)=4x+2
P(x)-C(x)=20x-0.1x^2-4x-2=-0.1x^2+16x-2
P(x)=-0.1(x^2-16x)-2
complete the square
P(x)=-0.1(x^2-16x+64)+6.4-2
P(x)=-0.1(x-8)^2+4.4
This is an equation of a parabola that opens down with vertex at (8, 4.4)
maximum profit=4.4 when units sold=8