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.
  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
  1. 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. $y + 2x = 100$
  1. Area of the garden as a function of $x$, say $A(x),$ can be represented as.
  1. $200 + 2x^2$
  2. $x - 2x^2$
  3. $200x - 2x^2$
  4. $200 - x^2$
  1. Maximum value of $A(x)$ occurs at $x$ equals.
  1. $50$ ft
  2. $30$ ft
  3. $26$ ft
  4. $31$ ft
  1. Maxi mum area of garden will be.
  1. $2500$ sq. ft
  2. $4000$ sq. ft
  3. $5000$ sq. ft
  4. $6000$ sq. ft

Answer

  1. (b) Area
Solution:
To create a garden using $200$ ft fencing, we need to maximise its area.
  1. (c) $y + 2x = 200$
Solution:
Required relation is given by $2x + y = 200.$
  1. (c) $200x - 2x^2$
Solution:
Area of garden as a function of x can be rep resented as
$A(x) = x·y = x(200 - 2x) = 200x - 2x^2$
  1. (a) $50$ ft
Solution:
$A(x) = 200x - 2x^2 \Rightarrow A'(x) = 200 - 4x$
For the area to be maximum $A'(x) = 0$
$⇒ 200 - 4x =0$
$⇒ x = 50$ ft
  1. (c) $5000$ sq. ft
Solution:
Maximum area of the garden $= 200(50) - 2(50)^2 = 10000 - 5000 = 5000$ sq. ft

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