Shobhit's father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in figure. He has 200 ft of wire fencing. Based on the above information, answer the following questions. 1. To construct a garden using 200 ft of fencing, we need to maximise its. 1. Volume 2. Area 3. Perimeter 4. Length of the side 2. If x denote the length of side of garden perpendicular to brick wall and y denote the length of side parallel to brick wall, then find the relation representing total amount of fencing wire. 1. x + 2y = 150 2. x + 2y = 50 3. y + 2x = 200 4. y + 2x = 100 3. Area of the garden as a function of x, say A(x), can be represented as. 1. 200 + 2x2 2. x - 2x2 3. 200x - 2x2 4. 200 - x2 4. Maximum value of A(x) occurs at x equals. 1. 50 ft 2. 30 ft 3. 26 ft 4. 31 ft 5. Maxi mum area of garden will be. 1. 2500 sq. ft 2. 4000 sq. ft 3. 5000 sq. ft 4. 6000 sq. ft

Shobhit's father wants to construct a rectangular garden using a …

Question

Shobhit's father wants to construct a rectangular garden using a brick wall on one side of the garden and wire fencing for the other three sides as shown in figure. He has 200 ft of wire fencing.

Based on the above information, answer the following questions.

To construct a garden using 200 ft of fencing, we need to maximise its.

Volume
Area
Perimeter
Length of the side

If x denote the length of side of garden perpendicular to brick wall and y denote the length of side parallel to brick wall, then find the relation representing total amount of fencing wire.

x + 2y = 150
x + 2y = 50
y + 2x = 200
y + 2x = 100

Area of the garden as a function of x, say A(x), can be represented as.

200 + 2x²
x - 2x²
200x - 2x²
200 - x²

Maximum value of A(x) occurs at x equals.

50 ft
30 ft
26 ft
31 ft

Maxi mum area of garden will be.

2500 sq. ft
4000 sq. ft
5000 sq. ft
6000 sq. ft

✓

Answer

(b) Area

Solution:

To create a garden using 200 ft fencing, we need to maximise its area.

Solution:

Required relation is given by 2x + y = 200.

Solution:

Area of garden as a function of x can be rep resented as

A(x) = x·y = x(200 - 2x) = 200x - 2x²

(a) 50 ft

Solution:

A(x) = 200x - 2x² ⇒ A'(x) = 200 - 4x

For the area to be maximum A'(x) = 0

⇒ 200 - 4x =0

⇒ x = 50 ft

Solution:

Maximum area of the garden = 200(50) - 2(50)² = 10000 - 5000 = 5000 sq. ft

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Application of Derivatives — Maths STD 12 Science — Question

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