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Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 8 maths

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Answer: C(x)=0.075 x^{2}+2 x+100

Given:  C^{\prime}(x)=\frac{d C}{d x}=2+0.15 x \text { and } C(0)=100

To find: Total cost function C(x)

Hint: Use integration and then find out the value of integral constant

Solution: Given that

        \begin{aligned} &=C^{\prime}(x)=\frac{d C}{d x}=2+0.15 x \\\\ &=d C=(2+0.15 x) d x \end{aligned}

Integrating on both sides we get,

        \begin{aligned} &=\int d C=\int(2+0.15 x) d x \\\\ &=>\int d C=\int 2 \mathrm{~d} x+0.15 \int x d x \\\\ &=>C=2 x+\frac{0.15 x^{2}}{2}+D \ldots(i) \end{aligned}

We have C=100 when x=0

        \begin{aligned} &=>100=2 \times 0+\frac{0.15 \times(0)^{2}}{2}+D \\ &=>D=100 \end{aligned}

Putting the value of D in equation (i) we get,

        \begin{aligned} &=>C=2 x+\frac{0.15 x^{2}}{2}+100 \\\\ &=>C(x)=2 x+0.075 x^{2}+100 \end{aligned}

Thus the total cost function C(x)=2 x+0.075 x^{2}+100

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