Question

consider the cost function \\(c(y) = \frac{y^3}{2} + 1\\)

(a) (5 points) calculate the average cost, average variable cost, average fixed cost, and marginal cost functions.
(b) (5 points) find the lowest points of the average variable cost and average cost curves.
(c) (5 points) draw all three cost curves on the same graph.

Explanation:

Identify cost components

The total cost function is given by:

$$ c(y) = \frac{y^3}{2} + 1 $$

The variable cost component is \(v(y) = \frac{y^3}{2}\) and the fixed cost component is \(F = 1\).

Calculate cost functions

The average cost (\(AC\)), average variable cost (\(AVC\)), average fixed cost (\(AFC\)), and marginal cost (\(MC\)) are:

$$ AC(y) = \frac{c(y)}{y} = \frac{y^2}{2} + \frac{1}{y} $$

$$ AVC(y) = \frac{v(y)}{y} = \frac{y^2}{2} $$

$$ AFC(y) = \frac{F}{y} = \frac{1}{y} $$

$$ MC(y) = \frac{dc(y)}{dy} = \frac{3y^2}{2} $$

Minimize average variable cost

To find the lowest point of \(AVC(y) = \frac{y^2}{2}\) for \(y \ge 0\):
The derivative is:

$$ \frac{dAVC}{dy} = y $$

Setting this to zero gives \(y = 0\).
The minimum value is:

$$ AVC(0) = 0 $$

Thus, the lowest point of the \(AVC\) curve is at \((0, 0)\).

Minimize average cost

To find the lowest point of \(AC(y) = \frac{y^2}{2} + \frac{1}{y}\) for \(y > 0\):
The derivative is:

$$ \frac{dAC}{dy} = y - \frac{1}{y^2} $$

Setting this to zero:

$$ y - \frac{1}{y^2} = 0 \implies y^3 = 1 \implies y = 1 $$

The minimum value is:

$$ AC(1) = \frac{1^2}{2} + \frac{1}{1} = 1.5 $$

Thus, the lowest point of the \(AC\) curve is at \((1, 1.5)\).

Answer:

(a) The calculated cost functions are:

• Average Cost: \(AC(y) = \frac{y^2}{2} + \frac{1}{y}\)
• Average Variable Cost: \(AVC(y) = \frac{y^2}{2}\)
• Average Fixed Cost: \(AFC(y) = \frac{1}{y}\)
• Marginal Cost: \(MC(y) = \frac{3y^2}{2}\)

(b) The lowest points of the curves are:

• Lowest point of \(AVC\): \((0, 0)\)
• Lowest point of \(AC\): \((1, 1.5)\)

(c) The curves are plotted below.