Question

a rectangular field is 65 meters wide and 105 meters long. give the length and width of another rectangular field 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 \( P = 2\times (l + w) \), where \( l \) is the length and \( w \) is the width. For the original rectangle, \( l = 105 \) meters and \( w = 65 \) meters. So, \( P = 2\times(105 + 65)=2\times170 = 340 \) meters.

Step2: Find the sum of length and width for the new rectangle

Since the perimeter is the same, for the new rectangle, \( 2\times(l' + w')=340 \), so \( l' + w'=\frac{340}{2}=170 \) meters. We know that for a given perimeter, the area of a rectangle is maximized when it is a square (i.e., \( l' = w' \)). But we need a rectangle (not necessarily a square, but with \( l'
eq w' \) but closer to each other than 105 and 65) with the same perimeter and larger area. Let's try to make the length and width closer. The average of 105 and 65 is \( \frac{105 + 65}{2}=85 \). Let's take \( w' = 80 \) meters, then \( l'=170 - 80 = 90 \) meters. Let's check the area of the original rectangle: \( A = 105\times65 = 6825 \) square meters. The area of the new rectangle with \( l' = 90 \) and \( w' = 80 \) is \( 90\times80 = 7200 \) square meters, which is larger. (We can also take other values closer to 85, like 82 and 88, etc. Here we take 80 and 90 as an example.)

Answer:

width = 80 meters
length = 90 meters

(Note: There are multiple correct answers, for example, width = 75 meters and length = 95 meters, etc., as long as the sum of length and width is 170 and the area is larger than 6825. The key is that the length and width are closer to each other than 105 and 65.)