Michele H.

asked • 05/19/16

A rectangle yard to be enclosed w/160 ft of fence. The 4th side is the barn. Find the dimensions of the enclosure and the maximum area that can be enclosed.

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Jose R.

I'll do it again. H=40 so substitute this into the area equation from earlier. A(40)=-2(40)2+160(40)=3200. Therefore the maximum area to be enclosed is 3200ft. Now L=40 so substitute this in 160=2(40)+W and solve for W. The dimensions are L=40 and W=80. Hope this was as clear as possible and helpful.
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05/19/16

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Jose R. answered • 05/19/16

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Hi Michele, you're working with area and perimeter here. The perimeter is 160ft. The formula is P=2L+2W, where L Is the length and W is the width. One side of the yard is the barn so your perimeter formula will be P=2L +W. Substitute 160 for P to get 160= 2L+W. Solve for W now to get W=-2L+160. Area is equal to A=LW. Substitute W into the area formula and distribute to get A=-2L2 +160L. This the graph of a parabola facing down and the k value for the vertex (h,k) is where it's maximized. Use -b/2a to find h first. h=-160/2(-2)=40
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Jose R.

I didn't have more room to type. I will finish it here. h=40 which happens to be the length. Substitute this in your area formula from earlier to give you A(40)=-2(40)2+160(40)=3200ft. The maximum area to be enclosed is 3200ft. To find the width substitute L=40 into 160=2(40)+W and solve for W. W=80ft and L=40ft. Hope this helped.
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05/19/16

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