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Q.The relation between “Marginal Cost (MC)” and “Average Cost (AC)” of producing ‘x’ units of a product is : (A) ddx(AC)=x(MC−AC)\dfrac{d}{dx}(AC) = x(MC - AC) (B) ddx(AC)=x(AC−MC)\dfrac{d}{dx}(AC) = x(AC - MC) (C) ddx(AC)=1x(MC−AC)\dfrac{d}{dx}(AC) = \dfrac{1}{x}(MC - AC) (D) ddx(AC)=1x(AC−MC)\dfrac{d}{dx}(AC) = \dfrac{1}{x}(AC - MC)

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★
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Differentiating AC=C/xAC=C/x by the quotient rule and substituting MC=dCdxMC=\dfrac{dC}{dx} gives ddx(AC)=1x(MC−AC)\dfrac{d}{dx}(AC)=\dfrac{1}{x}(MC-AC).

Total cost C(x)C(x); MC=dCdxMC = \dfrac{dC}{dx}; AC=CxAC = \dfrac{C}{x}; quotient rule ddx ⁣(uv)=u′v−uv′v2\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right)=\dfrac{u'v-uv'}{v^2}.

  1. Write average cost: AC=CxAC = \dfrac{C}{x}.
  2. Differentiate using the quotient rule: ddx(AC)=x dCdx−Cx2\dfrac{d}{dx}(AC) = \dfrac{x\,\frac{dC}{dx} - C}{x^2}.
  3. Replace dCdx\frac{dC}{dx} with MCMC: ddx(AC)=x MC−Cx2\dfrac{d}{dx}(AC) = \dfrac{x\,MC - C}{x^2}. …

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