Question

a rectangular field is 85 yards wide and 135 yards long give the length and the width of another rectangular field that has the same perimeter but a smaller area

Answer 1

The given rectangle has length (l) 135 yards, and width (w) 85 yards.

So the perimeter (P) of this rectangle will be,

`\begin{gathered} P=2(l+w) \\ P=2(135+85) \\ P=2*220 \\ P=440 \end{gathered}`

And the corresponding area (A) will be,

`\begin{gathered} A=L* W \\ A=135*85 \\ A=11475 \end{gathered}`

Let there be another rectangle with length (L) and width (W).

It is given that both therectangle has the same perimeter,

`\begin{gathered} 2(L+W)=440 \\ L+W=220 \\ W=220-L \end{gathered}`

Also the area of this new rectangle should be less than the area of the initial rectangle,

`\begin{gathered} L* W\leq A \\ L*(220-L)\leq11475 \\ 220L-L^2\leq11475 \\ L^2-220L+11475\ge0 \end{gathered}`

Solve the given equation as,

`\begin{gathered} L^2-85L-135L+11475\ge0 \\ L(L-85)-135(L-85)\ge0 \\ (L-85)(L-135)\ge0 \end{gathered}`

The product of two factors is positive if either both factors are positive, or both are negative.

CASE-1 Both factors positive.

`\begin{gathered} L-85\ge0\text{ and }L-135\ge0 \\ L\ge85\text{ and }L\ge135 \\ L\ge135 \end{gathered}`

CASE-2 Both factors negative.

`\begin{gathered} L-85\leq0\text{ and }L-135\leq0 \\ L\leq85\text{ and }L\leq135 \\ L\leq85 \end{gathered}`

So the given conditions are satisfied when the length of the new rectangle is either less than 85, or greater than 135.

For example, the new rectangle can be 120 units long. Then the corresponding width, as per the given condition should be,

`\text{Width}=220-120=100`

So the width should be 100 units.

Therefore, rectangle with length 120 units, and width 100 units, will also satisfy the given conditions.

(Note that the soution is not unique. This is just one of the many possible values of length and width of the rectangle)