Larsen’s Sharpening Shop, specializes in sharpening saws. Terry Larsen has determined that the cost of operating the shop is given by the function:
C(x) = x^2 – 40x + 430, where x is the number of saws sharpened daily.
a. Write C(x) = x^2 - 40x + 430 = a(x-h)^2 + k, identify a, h, and k.
b. What is the number of saws Terry must sharpen each day in order to minimize the operational cost?
c. What is the cost under these ideal condition.
Exactly what did you not understand about my previous response?
Sorry that post was a double from last night. It wasn't posted after you answered. Thank you by the way!
a. To write the given function, C(x) = x^2 - 40x + 430, in the form a(x-h)^2 + k, we need to complete the square.
Let's start by rearranging the terms of the given function:
C(x) = x^2 - 40x + 430
Now, we can complete the square by adding and subtracting (40/2)^2 = (20)^2 = 400 within the parentheses:
C(x) = x^2 - 40x + 400 - 400 + 430
Simplifying the expression inside the parentheses:
C(x) = (x^2 - 40x + 400) - 400 + 430
Now, we can rewrite the first three terms as a perfect square:
C(x) = (x - 20)^2 - 400 + 430
Simplifying further:
C(x) = (x - 20)^2 + 30
Therefore, the function C(x) = x^2 - 40x + 430 can be written as C(x) = (x - 20)^2 + 30, where a = 1, h = 20, and k = 30.
b. To minimize the operational cost, we need to find the minimum value of the function C(x) = (x - 20)^2 + 30.
Since the minimum value of a quadratic function occurs at its vertex, we can determine the number of saws Terry must sharpen each day by finding the x-coordinate of the vertex.
The x-coordinate of the vertex is given by the formula: x = -b / (2a)
For the given function, a = 1 and b = -40:
x = -(-40) / (2 * 1)
x = 40 / 2
x = 20
Therefore, Terry must sharpen 20 saws each day in order to minimize the operational cost.
c. Now that we know the number of saws Terry must sharpen each day is 20 (from part b), we can substitute this value into the function C(x) = (x - 20)^2 + 30 to find the cost under these ideal conditions.
C(20) = (20 - 20)^2 + 30
C(20) = (0)^2 + 30
C(20) = 30
Therefore, under these ideal conditions, the cost would be 30.