Question

A farmer is building a fence to enclose a rectangular area consisting of two separate regions. The four walls and one additional vertical segment (to separate the regions) are made up of fencing, as shown below. If the farmer has 162 feet of fencing, what are the dimensions of the region which enclose the maximal area?

Answer 1

Ich the Solved; your walls the rectangular area and the vertical segment are made up of fencing; Then The total =) (2L + 3W) feet Given that, the farmer has 162 feet of fencing; Thane fone; fence required 2L+3w=162 ore 2 2L= 162-3W L=81-3 W dangular area; Now, the area ag A=LxW Substituting eq. into equation 2 we will get; A= (81-3 W). W A = 81W- 3 w2 Taking desivative on the both sides of the above equation with respect to W, we will get; صدا

da 81 - 12 3 (24) = 81- 3W dw Now, I will be max. when T Therefore; 0=81-3W da dw =0 3W =81 W=27 Plugging this w=a7 into ea ① L= 81- 3 W L = 81- 31- 3 (27) L = 81- 40.5 L-40.S So, the Dimensions of the region which enclose max. area will be Length 40.5 Geet. and width = 27 geet. (Answer