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Q.The relation between 'Marginal cost' and 'Average cost' of producing 'x' units of a product is :

(a) d(AC)dx=x(MC−AC)\dfrac{d(AC)}{dx} = x(MC - AC)
(b) d(AC)dx=x(AC−MC)\dfrac{d(AC)}{dx} = x(AC - MC)
(c) d(AC)dx=1x(AC−MC)\dfrac{d(AC)}{dx} = \dfrac{1}{x}(AC - MC)
(d) d(AC)dx=1x(MC−AC)\dfrac{d(AC)}{dx} = \dfrac{1}{x}(MC - AC)
CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★
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Differentiating total cost C=x⋅ACC=x\cdot AC gives MC=AC+x d(AC)dxMC=AC+x\,\dfrac{d(AC)}{dx}, i.e. d(AC)dx=1x(MC−AC)\dfrac{d(AC)}{dx}=\dfrac{1}{x}(MC-AC).

Average cost AC=C(x)xAC=\dfrac{C(x)}{x} and marginal cost MC=dCdxMC=\dfrac{dC}{dx}, where C(x)C(x) is total cost of xx units.

  1. Write total cost as C=x⋅ACC=x\cdot AC.
  2. Differentiate w.r.t. xx (product rule): dCdx=AC+xd(AC)dx\dfrac{dC}{dx}=AC+x\dfrac{d(AC)}{dx}, i.e. MC=AC+xd(AC)dxMC=AC+x\dfrac{d(AC)}{dx}. …

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