Problema 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 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 provided by our tutors
R(x) = 20x-0.1x^2
C(x) = 4x+2
we plug the above values into P(x)=R(x)-C(x):
P(x) = (20x-0.1x^2) - (4x+2)
P(x) = -0.1x^2 + 16x - 2
we need to find the maximum of the quadratic function P(x) = -0.1x^2 + 16x - 2
since the quotient in front of x^2 is -0.1 < 0 the function has maximum
P max = c - (b^2)/(4a), where a = -0.1, b = 16, c = -2
P max = -2 - (16^2)/(4*(-0.1))
P max = $638
the maximum profit is $638.
lets find x for which P(x) = 638
x = -b/(2a)
x = - 16/(2*(-0.1))
x = 80 units
the number of units that must be sold to yield the maximum profit is 80 units.