Class 12 Applications Of Derivatives · Very Long Answer5 Marksconstrained_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #107

A 34 m wire is cut into two pieces, one forming a square and the other a rectangle whose length is twice its breadth. Find the piece lengths that minimize their combined area.

✅ Step-by-Step Solution

Final Answer: 16 m for the square and 18 m for the rectangle.

Let p be the square-piece length; its side is p/4. The rectangle perimeter is 34−p, so its breadth is (34−p)/6 and length (34−p)/3, with 0<p<34.
The combined area is A=p²/16+(34−p)²/18.
A′=p/8−(34−p)/9=0 gives 9p=8(34−p), hence 17p=272 and p=16.
A″=1/8+1/9=17/72>0, proving the unique global minimum.
Use 16 m for the square (side 4 m) and 18 m for the rectangle (6 m×3 m).

📋 CBSE Stepwise Marking Scheme

[1 mark] Let p be the square-piece length; its side is p/4. The rectangle perimeter is 34−p, so its breadth is (34−p)/6 and length (34−p)/3, with 0<p<34.
[1 mark] The combined area is A=p²/16+(34−p)²/18.
[1 mark] A′=p/8−(34−p)/9=0 gives 9p=8(34−p), hence 17p=272 and p=16.
[1 mark] A″=1/8+1/9=17/72>0, proving the unique global minimum.
[1 mark] Use 16 m for the square (side 4 m) and 18 m for the rectangle (6 m×3 m).

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Class 12 Maths · Applications Of Derivatives · Very Long Answer #107

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 6: Applications Of Derivatives

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