Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 37.

Answers (1)

240 items

Hint: For maximum or minimum value of P, we must have \frac{dP}{dx}=0

Given: Profit = S.P - C.P

P= x\left ( 5-\frac{x}{100} \right )-\left ( 500+\frac{x}{5} \right )

Solution: P= x\left ( 5-\frac{x}{100} \right )-\left ( 500+\frac{x}{5} \right )

\begin{aligned} &P=5 x-\frac{x^{2}}{100}-500-\frac{x}{5} \\ &\frac{d P}{d x}=5-\frac{x}{50}-\frac{1}{5} \\ &\frac{d P}{d x}=0 \\ &5-\frac{x}{50}-\frac{1}{5}=0 \\ &\frac{24}{5}=\frac{x}{50}, x=\frac{24 \times 50}{5}=240 \end{aligned}

Now,

\frac{d^2P}{dx^2}=\frac{-1}{50}<0

Profit is maximum if 240 items are sold.

 

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads