Question:
Consider the short-run cost function $C(q)=10q^{3}-80q^{2}+300q+50$. At the minimum average variable cost (AVC), the value of marginal cost (MC) is ____________ (in integer).
Consider the short-run cost function $C(q)=10q^{3}-80q^{2}+300q+50$. At the minimum average variable cost (AVC), the value of marginal cost (MC) is ____________ (in integer).
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For cubic $C(q)$ with a fixed constant, minimize $AVC=\frac{VC}{q}$ via derivative; at that $q$, $MC$ intersects $AVC$.
Updated On: Sep 1, 2025
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Correct Answer: 140
Solution and Explanation
Step 1: Identify VC and AVC.
$VC=10q^{3}-80q^{2}+300q \;\Rightarrow\; AVC=\dfrac{VC}{q}=10q^{2}-80q+300.$
Step 2: Find $q$ at minimum AVC.
$\dfrac{d(AVC)}{dq}=20q-80=0 \Rightarrow q=4.$
Step 3: Use $MC$ at $q=4$.
$MC=\dfrac{dC}{dq}=30q^{2}-160q+300 \Rightarrow MC(4)=30(16)-160(4)+300=140.$
(Also, at the minimum of AVC, $MC=AVC=10(16)-80(4)+300=140$.)
$VC=10q^{3}-80q^{2}+300q \;\Rightarrow\; AVC=\dfrac{VC}{q}=10q^{2}-80q+300.$
Step 2: Find $q$ at minimum AVC.
$\dfrac{d(AVC)}{dq}=20q-80=0 \Rightarrow q=4.$
Step 3: Use $MC$ at $q=4$.
$MC=\dfrac{dC}{dq}=30q^{2}-160q+300 \Rightarrow MC(4)=30(16)-160(4)+300=140.$
(Also, at the minimum of AVC, $MC=AVC=10(16)-80(4)+300=140$.)
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