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#School
#CBSE 12 Class
#R.D. Sharma
#RD Sharma Vol. 1 Class 12
#Maxima and Minima
#Maths

#1.3
#9.12
#Fill in the blanks
#Very Short Answer type
#Multiple Choice Questions (MCQs)
#17.1
#17.2
#17.3
#17.4
#17.5
#Case Study Based Questions (CSBQ)

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Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 1subquestion (ii) Maths Textbook Solution.

Answers (1)

Answer: $\frac{2}{\pi}\left ( 100x-x^2 \right )$

Hint: Area of rectangle = length × breadth

Solution:

Area =Area of rectangle

$=x\times y$

$\\ since, 2x+ \pi y =200\$

$\ \pi y =200-2x\$

$y =\frac{1}{\pi}(200-2x)$

Area = $x\times \frac{1}{\pi}(200-2x)$

Area = $\frac{1}{\pi} \left ( 200x-2x^{2} \right )$

$=\frac{2}{\pi} \left ( 100x-x^{2}\right )$

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Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 1subquestion (ii) Maths Textbook Solution.

Answers (1)

Posted by

infoexpert27

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