CBSE Class 12 Mathematics 2026 Supplementary Paper 65(B)/7 · Q374 Marksoptimization
CBSE Class 12 Mathematics 2026 Supplementary Paper 65(B)/7 · Question 37 (Applications Of Derivatives)
A rectangular garden uses a house wall as one side. There are 300 m of fencing for the other three sides. Let x be each side perpendicular to the wall and y the side parallel to it.
Part (1) 1 Marks
Write the fencing constraint.
Answer: 2x+y=300.
- The two perpendicular sides each use x metres.
- The remaining fenced side uses y metres, so the total is 2x+y.
Part (2) 1 Marks
Express the area in terms of x.
Answer: A(x)=300x−2x², for 0<x<150.
- Solve the constraint as y=300−2x.
- Multiply xy to obtain 300x−2x².
Part (3) 2 Marks
Find x for maximum area.
Answer: 75 m.
- Differentiate A to get 300−4x.
- The critical point is x=75 and A″=−4<0.
- Thus the maximum is at x=75 m.
Part (4) 2 Marks
Find the maximum area.
Answer: 11250 m².
- Complete the square: A(x)=11250−2(x−75)².
- The negative squared term cannot exceed zero.
- It vanishes at x=75, giving maximum area 11250 m².