The perfect pizza parlor estimates the average daily cost per pizza to be C(x) = (0.00025x^2 + 8x + 10)/x , where x is a number of pizzas made in a day.

a) determine the total cost at the level of production of 50 pizzas a day.

b) determine the production level that would minimize the average daily cost per pizza.

c) what is the minimum average daily cost per pizza?

cost per pizza at x = 50 is

[.00025(50)^2 + 8(50) +10 ]/50 = 8.21
* 50 = 410.63

dc/dx = [x (.0005 x + 8) -.00025x^2-8x-10) ]/x^2
when is the numerator zero?
.0005 x^2 + 8 x -.00025 x^2 -8 x -10 = 0
.00025 x^2 = 10
x = 200

for c, use 200 for x

(a) Plug in 50 for x and compute C(x).

(b) Solve for the value of x for which
C'(x) = dC(x)/dx = = 0.00025 x -20/x
= 0
25x^2 = 20,000

(c) Compute C(x) using the value of x from (b)

thanks :)

a) To determine the total cost at the level of production of 50 pizzas a day, we need to substitute x = 50 into the cost function C(x) = (0.00025x^2 + 8x + 10)/x.

C(50) = (0.00025(50)^2 + 8(50) + 10)/50
C(50) = (0.00025(2500) + 400 + 10)/50
C(50) = (0.625 + 400 + 10)/50
C(50) = 410.625/50
C(50) = 8.2125

Therefore, the total cost at the level of production of 50 pizzas a day is approximately $8.21.

b) To determine the production level that would minimize the average daily cost per pizza, we need to find the value of x at which the derivative of the cost function C(x) is equal to zero.

C(x) = (0.00025x^2 + 8x + 10)/x

To find the derivative, we can use the quotient rule:

C'(x) = [(0.00025x^2 + 8x + 10)(1) - (0.00025x^2 + 8x + 10)(1)]/x^2
C'(x) = (0.00025x^2 + 8x + 10 - 0.00025x^2 - 8x - 10)/x^2
C'(x) = (0)/x^2
C'(x) = 0

Therefore, the derivative of the average daily cost function is always zero. This means that the average daily cost function is constant and has no minimum or maximum.