Question

a rectangular lot is 125 meters long and 75 meters wide. give the length and width of another rectangular lot that has the same perimeter but a larger area. width = meters length = meters

Explanation:

Step1: Calculate the perimeter of the original rectangle

The formula for the perimeter of a rectangle is \(P = 2(l + w)\). Given \(l = 125\) meters and \(w=75\) meters.

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Step2: Express the relationship between length and width for the new rectangle

Let the length of the new rectangle be \(x\) meters and the width be \(y\) meters. Since \(P = 2(x + y)=400\), then \(x + y=200\), so \(y = 200 - x\).
The area of the new rectangle \(A=xy=x(200 - x)=200x-x^{2}\). The area of the original rectangle \(A_{0}=125\times75 = 9375\) square - meters.
We know that for a quadratic function \(A(x)=-x^{2}+200x\), its vertex occurs at \(x=-\frac{b}{2a}\) (for \(y = ax^{2}+bx + c\), here \(a=-1\), \(b = 200\)). The vertex \(x = 100\) (by \(x=-\frac{200}{2\times(-1)}=100\)), and \(A(100)=100\times(200 - 100)=10000\) (a square has the maximum area for a given perimeter among rectangles). But we can also choose non - square values. For example, if \(x = 110\)

Step3: Find the width when length is 110

Since \(x + y=200\), when \(x = 110\), then \(y=200 - 110=90\)
The area of the rectangle with \(x = 110\) and \(y = 90\) is \(A=110\times90 = 9900\)
The perimeter of the rectangle with \(l = 110\) and \(w = 90\) is \(P=2(110 + 90)=2\times200 = 400\)

Answer:

width \(=90\) meters, length \(=110\) meters