3q2+5q+300
Find the marginal and average cost functions.
Show that AC is at its minimum when
q = 10 and that MC = AC at this output
Marginal cost is the first derivitive of the total cost function. So,,
MC = 6q + 5
AC = 3q + 5 + 300/q. To find the minimum AC, take the first dirivitive and set equal to zero.
AC' = 3 - 300/q^2 = 0. Solve for q. Thus, q=10 at the minimum of the AC curve. So, at q=10 AC = 30+5+30 = 65
and MC = 60+5 = 65
QED
To find the marginal cost (MC) and average cost (AC) functions, we need to differentiate the given cost function with respect to q.
Let's start with the given cost function:
C(q) = 3q^2 + 5q + 300
Taking the derivative of C(q) with respect to q gives us the marginal cost function (MC):
MC(q) = dC(q)/dq
MC(q) = d(3q^2 + 5q + 300)/dq
MC(q) = 6q + 5
To find the average cost function (AC), we divide the cost function C(q) by the quantity q:
AC(q) = C(q)/q
AC(q) = (3q^2 + 5q + 300)/q
AC(q) = 3q + 5 + 300/q
Now, let's determine the output (q) at which the average cost (AC) is at its minimum.
To find the minimum, we need to find the derivative of the average cost (AC) function with respect to q, and then set it equal to zero:
dAC(q)/dq = 0
d(3q + 5 + 300/q)/dq = 0
Simplifying, we get:
3 - 300/q^2 = 0
Let's solve for q:
300/q^2 = 3
q^2 = 100
q = ±10
Since we are considering a positive quantity, q = 10.
Therefore, when q = 10, the average cost (AC) is at its minimum.
To show that MC = AC at this output, let's substitute q = 10 into both the MC and AC functions:
MC(10) = 6(10) + 5
MC(10) = 65
AC(10) = 3(10) + 5 + 300/(10)
AC(10) = 35 + 5 + 30
AC(10) = 70
We can see that MC(10) = 65 and AC(10) = 70. Therefore, MC = AC at this output (q = 10).