Question

Building sheep pens : darsh and darpan are fencing off a large rectangular area to build some temporary holding pens. to prep the males, females, and kids, they are separated into three smaller and equal-size pens partitioned page 30 sample paper 24 cbse maths basic x click here to purchase hard books of cbse online sample papers. 20 sample paper in each subject and rs 500/- for 4 subjects click here download india’s best study app having pdfs of all sample paper and question bank within the large rectangle. (i) if 384 ft of fencing is available and the maximum area is desired, what will be the dimensions of the larger, outer rectangle? (ii) what will be the dimensions of the smaller holding pens?

Answer 1

Since a maximum area is desired, the dimensions of the larger, outer rectangle would be equal to 96 ft and 96 ft.

How to calculate the area of a rectangle?

Mathematically, the area of a rectangle is calculated by using this formula:

A = LW

Where:

• L is the length.

• W is the width.

• A is the area.

Since the perimeter shouldn't be larger than the available fencing, we have:

384 = 2L + 2W

2W = 384 - 2L

W = 192 - L

Now, the area becomes:

A = L × (192 - L)

A = 192L - L²

By using completing the square to find the roots, we have:

A = 192L - L²

0 = 192L - L² + 96² - 96²

0 = 96² - (L² - 192L + 96²)

0 = 9216 - (L - 96)²

L = 96 ft.

W = 192 - 96

W = 96 ft.

The dimensions of the smaller holding pens.

For the width of the smaller holding pens, we have:

Width = 96/3

Width = 31 ft.

However, the length of the smaller holding pens would remain the same as 96 feet.

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