jvillShyahsra7e jvillShyahsra7e
  • 02-05-2017
  • Mathematics
contestada

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.

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keuticus94 keuticus94
  • 02-05-2017
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
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