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Find the dimensions of a rectangle (in m) with perimeter 68 m whose area is as large as possible. (enter the dimensions as a comma-separated list.)

Sagot :

The dmension of the rectangle with perimeter of 68 m has the largest possible area as 17m by 17m.

Perimeter of a rectangle

perimeter of rectangle = 2(l + w)

where

  • l = length
  • w = width

Therefore,

perimeter = 68 m

68 = 2l + 2w

l + w = 34

l = 34 - w

Hence, the largest possible area is at the amximum.

area = lw

area = w(34 - w)

area = 34w - w²

The maximum is at dA / dw = 0

Therefore,

34 - 2w  = 0

2w = 34

w = 34 / 2

w = 17

l + w = 34

l + 17 = 34

l = 34 - 17

l = 17

area = 17² = 289 m²

learn more on rectangle here:https://brainly.com/question/17076667

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