Question

Find the dimensions of a rectangle with area 2,197 m2 whose perimeter is as small as possible. (if both values are the same number, enter it into both blanks.

Answer 1

Area of a rectangle A = l×w = 2197
l - length
w-width
l×w = 2197
making l the subject
l = $\frac{2197}{w}$
The perimeter of rectangle
p = 2(l+w)
   = 2(w+$\frac{2197}{w}$)
   =2w+$\frac{2(2197)}{w}$
To obtain minimal perimeter
$\frac{dp}{dw}=0$
2-$\frac{4394}{w^{2} }$ = 0
multiplying both sides of equation by $w^{2}$
2$w^{2}$-4394=0
2$w^{2}$=4394
$w^{2}$=2197
w = √2197
   w1 = 46.87
   w2 = 46.87