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May 16, 2026

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Actually,
P(x) = \frac{5000}{x} + 120 - 0.5x
P'(x) = -\frac{5000}{x^2} - 0.5 \quad \text{— still negative?}
Wait — no, derivative of \( -0.5x \) is \( -0.5 \), derivative of \( 5...
This is always negative, meaning cost per unit decreases as \( x \) in...
Wait — cost per unit: \( P(x) = \frac{5000}{x} + 120 - 0.5x \)
But as \( x \to 0^+ \), \( 5000/x \to \infty \); as \( x \to \infty \)...
Likely a typo in interpretation. For realistic models, average cost ty...
But given: \( C(x) = 5000 + 120x - 0.5x^2 \). This implies fixed start...
But proceeding with given function.

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