The profit is given by the subtraction of revenue and cost. So, the function for the profit is
[tex] P(x) = R(x) - C(x) = 60x-0.1x^2 - (6x+15) = 60x-\frac{1}{10}x^2 - 6x-15 = -\frac{1}{10}x^2 + 54x - 15 [/tex]
We want to maximize this function, so let's derive its expression and set it to zero:
[tex] P'(x) = -\frac{1}{5}x + 54 = 0 \iff -\frac{1}{5}x = -54 \iff x = 270 [/tex]
For this number of units, the profit equals
[tex] P(270) = -\frac{1}{10}(270)^2 + 54\cdot 270 - 15 = 7275 [/tex]
The number of units that must be produced and sold in order to yield the maximum profit is 270 units and the maximum profit made is $7275
The profit made is the difference between revenue and cost as shown below:
P(x) = R(x) - C(x)
Given the following functions
R(x)=60x-0.1x^2
C(x)=6x+15
Substitute
P(x) = 60x-0.1x^2 - 6x - 15
P(x) = -0.1x^2 + 54x - 15
The profit is maximized at when P'(x) = 0
P'(x) = -0.2x + 54
0.2x = 54
x = 54/0.2
x = 270
The number of units that must be produced and sold in order to yield the maximum profit is 270 units
Determine the maximum profit
P(270) = -0.1(270)^2 + 54(270) - 15
P(270) = $7275
Hence the maximum profit made is $7275
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