A steel plant is capable of producing x tonnes per day of low-grade steel and y tonnes per day of a high-grade steel, where $y=\frac{32 – 5x}{10-x}$. If the fixed price of low-grade steel is half that of high-grade steel, find the quantity of low-grade steel that should be produced per day for maximum receipts.

4 tonnes

The correct answer is Option (2) → 4 tonnes

Let ₹p per tonne be the fixed market price of low-grade steel, then ₹2p per tonne is the fixed market price of high-grade steel. Let R be the total revenue, then

$R = px + 2py = px + 2p.\frac{32 – 5x}{10- x}= p.\frac{- x^2 + 64}{10-x}$

$∴\frac{dR}{dx}=p.\frac{(10-x)(-2x)-(-x^2+64)(-1)}{(10 - x)^2}=p.\frac{x^2-20x+64}{(10 - x)^2}$

For R to be maximum, $\frac{dR}{dx}=0$

$⇒ x^2 - 20x + 64 = 0 ⇒ (x-4)(x - 16) = 0⇒ x= 4, 16$.

Now, $\frac{dR}{dx}= p\frac{x^2-20x+64}{(10 - x)^2}=\frac{(10 - x)^2-36}{(10 - x)^2}=p\left(1-\frac{36}{(10 - x)^2}\right)$

$⇒ \frac{d^2R}{dx^2}=p(0 - 36.(-2)(10 - x)^{-3}.(-1)) = -\frac{72p}{(10 - x)^3}$.

$∴\left.\frac{d^2R}{dx^2}\right|_{x=4}=-\frac{72p}{6^3}=-\frac{p}{3}<0;\left.\frac{d^2R}{dx^2}\right|_{x=16}=-\frac{72p}{(-6)^3}=\frac{p}{3}>0$.

∴ R is maximum when $x = 4$.

Hence, the receipts are maximum when the steel plant produces 4 tonnes of low-grade steel per day.