Find the total profit, the maximum value of the total profit, and the value of x at which it occurs when the revenue is given by R(x)=240x-x^2 and the cost is given by C(x)=5000+8x
The total profit, P(x)=-x^2+_______x-5000
p = r - c
= 240 x - x^2 - 5000 - 8 x
= - x^2 + 232 x - 5000
x^2 - 232 x = -p - 5000
x^2 - 232 x + 13456 = -p - 5000 + 13456
(x - 116)^2 = -(p - 8456)
vertex at
profit = 8456
and
x = 116
To find the total profit, P(x), we subtract the cost, C(x), from the revenue, R(x). So, we have:
P(x) = R(x) - C(x)
Given that the revenue function, R(x), is 240x - x^2, and the cost function, C(x), is 5000 + 8x, we can substitute these values into the equation for P(x):
P(x) = (240x - x^2) - (5000 + 8x)
Simplifying further,
P(x) = 240x - x^2 - 5000 - 8x
Combining like terms,
P(x) = -x^2 + (240x - 8x) - 5000
P(x) = -x^2 + 232x - 5000
So, the total profit function, P(x), is given by -x^2 + 232x - 5000.
To find the maximum value of the total profit, we can use calculus. The maximum or minimum point of a quadratic function occurs at its vertex, which is given by x = -b/(2a), where a and b are the coefficients of x^2 and x, respectively. In our case, a = -1 and b = 232.
Using the formula, we can find the value of x at which the maximum profit occurs:
x = -232 / (2 * -1)
x = -232 / -2
x = 116
Therefore, the value of x at which the maximum profit occurs is 116.
To find the maximum profit value, we substitute this value of x into the total profit function:
P(x) = -x^2 + 232x - 5000
P(116) = -(116)^2 + 232(116) - 5000
P(116) = -13456 + 26912 - 5000
P(116) = 8424
So, the maximum value of the total profit is 8424.