A steel plant is capable of producing ‘x’ tonnes per day of low-grade steel and ‘y’ tonnes per day of high-grade steel, where \( y=\displaystyle\frac{32-5x}{10-x} \). If the fixed price of low-grade steel is half that of high-grade steel, then the quantity of low-grade steel that should be produced per day for maximum receipts is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4
Let the price of low-grade steel be Rs. p per tonne; then high-grade steel sells at Rs. 2p per tonne. The daily receipts are
\( \displaystyle R=px+2py \)
\( \displaystyle =p\left[x+\frac{2(32-5x)}{10-x}\right] \)
\( \displaystyle =p\left[x+\frac{64-10x}{10-x}\right] \)
By the quotient rule,
\( \displaystyle \frac{d}{dx}\left(\frac{64-10x}{10-x}\right)=\frac{(10-x)(-10)-(64-10x)(-1)}{(10-x)^2} \)
\( \displaystyle =\frac{-100+10x+64-10x}{(10-x)^2} \)
\( \displaystyle =\frac{-36}{(10-x)^2} \)
\( \displaystyle \frac{dR}{dx}=p\left[1-\frac{36}{(10-x)^2}\right]=0 \)
\( \displaystyle (10-x)^2=36 \)
\( \displaystyle 10-x=\pm 6 \)
\( \displaystyle x=4 \text{ or } x=16 \)
\( \displaystyle \frac{d^2R}{dx^2}=p\cdot\frac{-72}{(10-x)^3} \)
At x = 4, \( (10-x)^3=216\gt 0 \), so \( \frac{d^2R}{dx^2}\lt 0 \): a maximum. At x = 16, \( (10-x)^3=-216 \), so it is a minimum.
Hence 4 tonnes of low-grade steel should be produced per day.
Why the other options are incorrect
Option (2) 16 — Incorrect. x = 16 also makes \( \frac{dR}{dx}=0 \), but there \( \frac{d^2R}{dx^2}\gt 0 \), so the receipts are a minimum, not a maximum.
Option (3) 20 — Incorrect. At x = 20, \( (10-x)^2=100\ne 36 \), so \( \frac{dR}{dx}\ne 0 \).
Option (4) 12 — Incorrect. At x = 12, \( (10-x)^2=4\ne 36 \), so \( \frac{dR}{dx}\ne 0 \).