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May 05, 2026

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So assume \( C(x) = 5000 + 120x + 0.5x^2 \). Then average cost:
P(x) = \frac{5000}{x} + 120 + 0.5x
Now take derivative:
P'(x) = -\frac{5000}{x^2} + 0.5
-\frac{5000}{x^2} + 0.5 = 0 \Rightarrow \frac{5000}{x^2} = 0.5 \Righta...
Since \( x > 0 \), \( x = 100 \).
Second derivative: \( P''(x) = \frac{10000}{x^3} > 0 \) for \( x > 0 \...
Thus, production level 100 units minimizes average cost.
Even though original function has \( -0.5x^2 \), which breaks realism,...
Given the instruction to generate similar difficulty, we proceed with...

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