Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #78

A window is a rectangle surmounted by an equilateral triangle sharing its top edge. Its exterior perimeter is 12 m. Find the rectangle dimensions that maximize the whole window area.

✅ Step-by-Step Solution

Final Answer: Width 12/(6−√3) m; height (18−6√3)/(6−√3) m.

Let the rectangle width and height be x,y>0. The triangle has side x, so the exterior perimeter gives 3x+2y=12 and y=(12−3x)/2, 0<x<4.
The total area is A=xy+√3x²/4=6x−3x²/2+√3x²/4.
A′=6−3x+√3x/2=0 gives x=12/(6−√3).
A″=−3+√3/2<0 throughout the feasible interval, so this interior point is the global maximum.
The rectangle height is y=(12−3x)/2=(18−6√3)/(6−√3) m, with the stated width.

📋 CBSE Stepwise Marking Scheme

[1 mark] Let the rectangle width and height be x,y>0. The triangle has side x, so the exterior perimeter gives 3x+2y=12 and y=(12−3x)/2, 0<x<4.
[1 mark] The total area is A=xy+√3x²/4=6x−3x²/2+√3x²/4.
[1 mark] A′=6−3x+√3x/2=0 gives x=12/(6−√3).
[1 mark] A″=−3+√3/2<0 throughout the feasible interval, so this interior point is the global maximum.
[1 mark] The rectangle height is y=(12−3x)/2=(18−6√3)/(6−√3) m, with the stated width.

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Class 12 Maths · Applications Of Derivatives · Very Long Answer #78

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 6: Applications Of Derivatives

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