Let's work through each part of the problem step by step.
Given Cost Function:
\[ C(x) = 54,000 + 240x + 4x^{3/2} \]
(a) Total Cost at 1000 Units
To find the total cost at a production level of 1,000 units, substitute \( x = 1000 \) into the cost function \( C(x) \): \[ C(1000) = 54,000 + 240(1000) + 4(1000)^{3/2} \] Calculating each term:
- \( 240(1000) = 240,000 \)
- \( (1000)^{3/2} = 1000 \times \sqrt{1000} = 1000 \times 31.6228 \approx 31,622.8 \)
- \( 4(1000)^{3/2} = 4 \times 31,622.8 \approx 126,491.2 \)
So, \[ C(1000) = 54,000 + 240,000 + 126,491.2 \approx 420,491.2 \]
Thus, the total cost at a production level of 1,000 units is: \[ \boxed{420491.20} \]
(b) Average Cost at 1000 Units
The average cost \( c(x) \) is given by: \[ c(x) = \frac{C(x)}{x} \] So, for \( x = 1000 \): \[ c(1000) = \frac{C(1000)}{1000} = \frac{420,491.2}{1000} = 420.4912 \]
Rounded to the nearest cent, the average cost is: \[ \boxed{420.49} \]
(c) Marginal Cost at 1000 Units
The marginal cost \( C'(x) \) is the derivative of \( C(x) \). First, we find \( C'(x) \): \[ C'(x) = 0 + 240 + 4 \cdot \frac{3}{2}x^{1/2} = 240 + 6x^{1/2} \] Now substitute \( x = 1000 \): \[ C'(1000) = 240 + 6(1000)^{1/2} = 240 + 6(31.6228) \approx 240 + 189.7368 \approx 429.7368 \]
Thus, the marginal cost at a production level of 1,000 units is: \[ \boxed{429.74} \]
(d) Production Level that Minimizes Average Cost
To find the production level that minimizes average cost, we need to set the derivative of the average cost \( c(x) \) to zero. First, the average cost is: \[ c(x) = \frac{C(x)}{x} = \frac{54,000 + 240x + 4x^{3/2}}{x} = \frac{54,000}{x} + 240 + 4x^{1/2} \] Find the derivative \( c'(x) \): \[ c'(x) = -\frac{54,000}{x^2} + 2x^{-1/2} \] Setting \( c'(x) = 0 \): \[ -\frac{54,000}{x^2} + 2x^{-1/2} = 0 \] Solving for \( x \): \[ 2x^{-1/2} = \frac{54,000}{x^2} \] Cross-multiplying: \[ 2x^{3/2} = 54,000 \] \[ x^{3/2} = 27,000 \] Taking both sides to the power of \( \frac{2}{3} \): \[ x = (27,000)^{2/3} = 30^2 = 900 \] So, the production level that minimizes average cost is: \[ \boxed{900} \]
(e) Minimum Average Cost
Now, find the average cost at \( x = 900 \): \[ C(900) = 54,000 + 240(900) + 4(900)^{3/2} \] Calculating each term:
- \( 240(900) = 216,000 \)
- \( (900)^{3/2} = 900 \times 30 = 27,000 \)
- \( 4(900)^{3/2} = 4 \times 27,000 = 108,000 \)
So, \[ C(900) = 54,000 + 216,000 + 108,000 = 378,000 \] Calculating the average cost: \[ c(900) = \frac{C(900)}{900} = \frac{378,000}{900} = 420 \] Thus, the minimum average cost is: \[ \boxed{420.00} \]
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