Answer:
L = 15 and W = 15. Hence, the dimensions of the rectangle with the smallest perimeter and area of 225 sq.ft. are 15 ft x 15 ft (i.e. a square).
Explanation:
Let's assume the length of the rectangle is L and the width is W. The area of the rectangle is given as 225 sq.ft. Therefore, we have:
L x W = 225
The perimeter of the rectangle is given by:
2L + 2W
We need to find the dimensions of the rectangle that will give the smallest possible perimeter. To do this, we need to minimize the expression for the perimeter subject to the constraint that the area is 225 sq.ft. We can use the method of Lagrange multipliers to solve this problem.
Let f(L, W) = 2L + 2W be the expression for the perimeter and g(L, W) = L x W - 225 be the expression for the area. We want to minimize f(L, W) subject to the constraint g(L, W) = 0.
The Lagrange function is given by:
L(L, W, λ) = f(L, W) - λg(L, W)
= 2L + 2W - λ(LW - 225)
Taking partial derivatives and setting them equal to zero, we get:
∂L/∂λ = W = 0
∂W/∂λ = L = 0
∂L/∂λ = -λW = 0
∂W/∂λ = -λL = 0
∂L/∂λ = 2 - Wλ = 0
∂W/∂λ = 2 - Lλ = 0
Solving these equations, we get:
λ = 2/L = 2/W
Substituting λ in the expression for the area, we get:
L^2 = 225
Therefore, L = 15 and W = 15. Hence, the dimensions of the rectangle with the smallest perimeter and area of 225 sq.ft. are 15 ft x 15 ft (i.e. a square).