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3. A firm’s average cost function is given by
AC = (800 : Q) +2Q= 18
(a) Find, to the nearest whole number, the value of Q at the lowest point on the graph of AC plotted against Q, in the interval, 0 ≤ Q ≤ 30.
(b) State the value of the fixed costs.

Respuesta :

MrRoyal

The value of the fixed cost is the product of the average cost function and the quantity.

  • The value of Q at the lowest point is 20
  • The value of the fixed cost is 1960

Given:

[tex]AC = \frac{800}{Q} + 2Q + 18[/tex]

Lowest value of Q

[tex]AC = \frac{800}{Q} + 2Q + 18[/tex]

Differentiate the average cost function

[tex]AC' = -800Q^{-2} + 2[/tex]

Equate to 0

[tex]-800Q^{-2} + 2 =0[/tex]

Collect like terms

[tex]- 800Q^{-2} =-2[/tex]

Rewrite as:

[tex]-\frac{800}{Q^2} = -2[/tex]

Solve for [tex]Q^2[/tex]

[tex]Q^2 = -\frac{800}{-2}[/tex]

[tex]Q^2 = 400[/tex]

Take square roots of both sides

[tex]Q= 20[/tex]

Fixed Cost

The fixed cost (FC) is: [tex]FC = AC \times Q[/tex]

So, we have:

[tex]FC = (\frac{800}{Q} + 2Q + 18) \times Q[/tex]

Substitute 20 for Q

[tex]FC = (\frac{800}{20} + 2\times 20 + 18) \times 20[/tex]

[tex]FC = (40 + 40 + 18) \times 20[/tex]

[tex]FC = 98 \times 20[/tex]

[tex]FC = 1960[/tex]

Hence, the value of the fixed cost is 1960

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