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3.4 · Q7

Q.The total cost function of a manufacturing company is given by C(x)=2x(x+4x+3)+3C(x) = 2x\left(\dfrac{x+4}{x+3}\right) + 3. Show that MC (Marginal Cost) falls continuously as the output 'x' increases.

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MC=C′(x)MC=C'(x) simplifies to 2+6(x+3)22+\dfrac{6}{(x+3)^2}, whose derivative −12(x+3)3-\dfrac{12}{(x+3)^3} is negative for every x≥0x\ge 0, proving MC decreases continuously.

Marginal Cost MC=dCdxMC=\dfrac{dC}{dx}. A quantity falls continuously if its derivative is negative throughout the domain. Quotient rule: (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}.

  1. Given C(x)=2x(x+4x+3)+3=2(x2+4x)x+3+3C(x)=2x\left(\dfrac{x+4}{x+3}\right)+3=\dfrac{2(x^2+4x)}{x+3}+3.
  2. Differentiate using the quotient rule (with u=x2+4x, v=x+3, u′=2x+4, v′=1u=x^2+4x,\ v=x+3,\ u'=2x+4,\ v'=1):

MC=C′(x)=2⋅(2x+4)(x+3)−(x2+4x)(1)(x+3)2.MC=C'(x)=2\cdot\dfrac{(2x+4)(x+3)-(x^2+4x)(1)}{(x+3)^2}.

  1. Expand the numerator: (2x+4)(x+3)=2x2+10x+12(2x+4)(x+3)=2x^2+10x+12; subtract (x2+4x)(x^2+4x) to get x2+6x+12x^2+6x+12.
  2. So MC=2(x2+6x+12)(x+3)2MC=\dfrac{2(x^2+6x+12)}{(x+3)^2}.
  3. Note x2+6x+12=(x+3)2+3x^2+6x+12=(x+3)^2+3, hence MC=2[(x+3)2+3](x+3)2=2+6(x+3)2.MC=\dfrac{2\big[(x+3)^2+3\big]}{(x+3)^2}=2+\dfrac{6}{(x+3)^2}. …

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