Question

A rectangular lot is 110 yard long and 60 yards wide. Give the length and width of another rectangular lot that has the same perimeter but larger area.

Answer 1

85 yards

Explanation:

Length = 110 yard

width = 60 yards

Perimeter of rectangle = 2 ( length + width)

P = 2 (110 + 60) = 340 yards

Now let the length is L and width is W.

P = 340 = 2 ( L + W)

L + W = 170 W = 170 - L ..... (1)

Area, A = L x W

A = L (170 - L)

A = 170 L - L²

Differentiate with respect to L

dA/dL = 170 - 2 L

Put it equal to zero for maxima and minima

2 L = 170

L = 85 yards

So, W = 170 - L = 170 - 85 = 85 yards

So, A = 85 x 85 = 7225 yard²

So, when the length and width is same and equal to 85 yards the perimeter is same and the area is largest.

Answer 2

Length and width of 85 yards will give the same perimeter and larger area.

Explanation:

Let x represent length and y represent width of rectangle.

The perimeter of the rectangle would be
`2x+2y\Rightarrow 2(x+y)`.

We have been given that a rectangular lot is 110 yard long and 60 yards wide. The perimeter of the given rectangle would be 2 times the width and length.

`\text{Perimeter}=2(110+60)`

`\text{Perimeter}=2(170)`

`\text{Perimeter}=340`

Upon equating both perimeters, we will get:

`2(x+y)=340`

Divide both sides by 2:

`x+y=170`

`y=170-x`

We know that area of rectangle is length times width.

`\text{Area}=x\cdot y`

`A(x)=x\cdot (170-x)`

`A(x)=170x-x^2`

Now, we will take the derivative of area function as:

`A'(x)=170-2x`

Now we will equate derivative with 0 and solve for x.

`0=170-2x`

`2x=170`

`(2x)/(2)=(170)/(2)`

`x=85`

Therefore, the length of rectangle would be 85 yards.

Upon substituting
`x=85` in equation
`y=170-x`, we will get:

`y=170-85=85`

Therefore, the width of rectangle would be 85 yards.

This means that we will get a square. Since each square is a rectangle, therefore, length and width of 85 yards will give the same perimeter and larger area.

We can verify our answers.

`\text{New area}=85* 85=7225`

`\text{Original area}=110\cdot 60=6600`

`\text{New perimeter}=2(85+85)`

`\text{New perimeter}=2(170)=340`