CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/2 · Q364 Markscase_study
CBSE Class 12 Mathematics 2026 Supplementary Paper 65/7/2 · Question 36 (Applications Of Derivatives)
A window consists of a rectangle of length x and breadth y metres, surmounted by an equilateral triangle on its length. Its outer boundary admits light through the whole opening.
Part (1) 1 Marks
For perimeter 18 m, write the relation between x and y.
Answer: 3x+2y=18.
- The outer boundary has the rectangle’s bottom x, two vertical sides y each, and two exposed triangle sides x each.
- The perimeter is 3x+2y, so 3x+2y=18.
Part (2) 1 Marks
With perimeter 18 m, express the window area as a function of x.
Answer: A(x)=9x−(3/2−√3/4)x², 0<x<6.
- The area is xy+√3x²/4 and y=(18−3x)/2.
- Substitution gives A(x)=9x−(3/2−√3/4)x² for positive x,y.
Part (3) 2 Marks
Find rectangle dimensions giving the largest total window area when its perimeter is 18 m.
Answer: x=18/(6−√3), y=(27−9√3)/(6−√3) metres.
- Differentiate the area: A′=9−(3−√3/2)x.
- Set it to zero to get x=18/(6−√3). Since A″=−3+√3/2<0 and 0<x<6, this is the interior maximum.
- Use y=9−3x/2 to obtain y=(27−9√3)/(6−√3).
Part (4) 2 Marks
If the total window area is 100 m², express its perimeter in terms of x.
Answer: P(x)=(3−√3/2)x+200/x.
- The area condition is xy+√3x²/4=100.
- Solve y=100/x−√3x/4 and substitute into P=3x+2y.
- Thus P=(3−√3/2)x+200/x for 0<x<20/3^(1/4), ensuring y>0.