#### Question

The total cost function of a firm is `C = x^2 + 75x + 1600` for output x. Find the output (x) for which average

cost is minimum. Is `C_A = C_M` at this output?

#### Solution

Given cost function

`C(x) = x^2 + 75x + 1600`

Average `bar C (x)=(C(x))/x`

=`(x^2+75x+1600)/x`

=`x+75+1600/x`

Now `barC'(x)=(dbarC(x))/dx=1-1600/x^2`

For minimum average cost `barC (x)=0`

∴Minimum average cost=`barC(x)=40+75+1600/40=155`

∴ `C_A=155`

Now we find marginal cost i.e.,

`C_m=(dC)/(Dx)`

C_m=`d/dx(x^2+75x+1600)`

= 2x + 75 ...(1)

∴ put x=40 in eq (1)

`C_m=2xx40+75`

= `80+75=155`

`C_A=C_m for x=40`

Is there an error in this question or solution?

#### APPEARS IN

Solution The Total Cost Function of a Firm is C = X 2 + 75 X + 1600 for Output X. Find the Output (X) for Which Average Cost is Minimum. is C a = C M at this Output? Concept: Maxima and Minima.