Question

a rectangular lot is 120 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 = \square meters length = \square meters

Explanation:

Step1: Calculate the perimeter of the original rectangle

The formula for the perimeter \( P \) of a rectangle is \( P = 2\times (l + w) \), where \( l \) is the length and \( w \) is the width. For the original rectangle, \( l = 120 \) meters and \( w = 75 \) meters. So, \( P = 2\times(120 + 75) = 2\times195 = 390 \) meters. The semi - perimeter (sum of length and width) is \( \frac{P}{2}=195 \) meters.

Step2: Recall the relationship between length, width and area of a rectangle

The area \( A \) of a rectangle is given by \( A=l\times w \), and since \( l + w = 195 \) (from the semi - perimeter), we can express \( l = 195 - w \). Then the area \( A=(195 - w)\times w=195w - w^{2} \). This is a quadratic function, and for a quadratic function \( y = ax^{2}+bx + c \) (here \( a=- 1\), \( b = 195 \), \( c = 0 \)), the maximum occurs at \( x=-\frac{b}{2a} \). In terms of \( w \), the width that maximizes the area is \( w=\frac{195}{2}=97.5 \) meters, and the corresponding length \( l = 195 - 97.5=97.5 \) meters (a square, since for a given perimeter, a square has the maximum area). But we can also choose other values where the length and width are closer to each other than 120 and 75. Let's pick a length and width such that \( l + w=195 \) and \( l>w \) (or vice - versa) and \( l\times w>120\times75 = 9000 \). Let's try \( l = 100 \) meters, then \( w=195 - 100 = 95 \) meters. The area is \( 100\times95 = 9500 \) square meters, which is larger than \( 120\times75=9000 \) square meters. (We can also use the square as the optimal case. The square with side length \( 97.5 \) meters has an area of \( 97.5\times97.5 = 9506.25 \) square meters which is larger than 9000. Here we will use the square values for the most optimal case.)

Step3: Determine the length and width

Since the semi - perimeter is 195 meters, if we take the rectangle as a square (which gives the maximum area for a given perimeter), the length and width are both \( \frac{390}{4}=97.5 \) meters. But we can also choose other non - square rectangles where length and width are closer. For example, if we take width \( = 90 \) meters, then length \(=195 - 90 = 105 \) meters. The area is \( 105\times90=9450 \) square meters which is larger than 9000. But the square gives the largest area. Let's use the square values.

Answer:

width \( = 97.5 \) meters, length \( = 97.5 \) meters (or other pairs like width = 95, length = 100; width = 90, length = 105 etc. as long as \( l + w = 195 \) and \( l\times w>9000 \))