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3.2 · Q5

Q.For what values of x is the rate of increase of total cost function C(x)=x3−5x2+5x+8C(x) = x^3 - 5x^2 + 5x + 8 is twice the rate of increase of x?

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The rate of increase of cost is twice that of xx means C′(x)=2C'(x)=2; solving 3x2−10x+5=23x^2-10x+5=2 gives x=13x=\frac{1}{3} or x=3x=3.

By the chain rule dCdt=C′(x)dxdt\dfrac{dC}{dt}=C'(x)\dfrac{dx}{dt}; the condition "rate of increase of CC is twice the rate of increase of xx" means dCdt=2dxdt\dfrac{dC}{dt}=2\dfrac{dx}{dt}, hence C′(x)=2C'(x)=2.

  1. Given cost function: C(x)=x3−5x2+5x+8.C(x)=x^3-5x^2+5x+8.

  2. Marginal (rate) function:

C′(x)=3x2−10x+5.C'(x)=3x^2-10x+5.

  1. Apply the condition C′(x)=2C'(x)=2:

3x2−10x+5=2.3x^2-10x+5=2.

  1. Form the quadratic: 3x2−10x+3=0.3x^2-10x+3=0. …

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