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