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Solve the following.
The cost function C(x) of producing 'x' quantities of a product is given by C(x) = 500x² + 2500x +…
The cost function $C(x)$ of producing '$x$' quantities of a product is given by $C(x) = 500x^2 + 2500x + 5000$. If each unit of the product is sold at ₹ 6000 then, calculate
(a)[2]
Break-even point
(b)[2]
Marginal cost and show that the marginal cost increases as the output '$x$' increases.
Answer
Answer (a)
Revenue $R(x)=6000x$. At break-even $R(x)=C(x)$:
$6000x=500x^2+2500x+5000$
$500x^2-3500x+5000=0$, i.e. $x^2-7x+10=0$
$(x-2)(x-5)=0$
Final answer: $x=2$ and $x=5$ units
Answer (b)
Marginal cost is the derivative of the cost function; it increases with $x$ if its own derivative is positive.
- 1. $C(x)=500x^2+2500x+5000$ (given).
- 2. Marginal cost $MC=\frac{dC}{dx}=1000x+2500$ (power rule).
- 3. $\frac{d(MC)}{dx}=\frac{d^2C}{dx^2}=1000$.
- 4. Since $\frac{d(MC)}{dx}=1000>0$ for all $x$, $MC$ is an increasing function of $x$.
- 5. Hence the marginal cost $1000x+2500$ increases as the output $x$ increases.