word problem in quadratic equation

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10-07-2014
Math

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word problem in quadratic equation

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jewelsonpp jewelsonpp · Answer 1 · 2014-07-30T15:35:50+08:00

The fencing-length information gives me perimeter. If the length of the enclosed area is L and the width is w, then the perimeter is 2L + 2w = 500, so L = 250 – w. By solving the perimeter equation for one of the variables, I can substitute into the area formula and get an equation with only one variable:A = Lw = (250 – w)w = 250w – w2 = –w2 + 250wTo find the maximum, I have to find the vertex (h, k).h = –b/2a = –(250)/2(–1) = –250/–2 = 125In my area equation, I plug in "width" values and get out "area" values. So the h-value in the vertex is the maximizing width, and the k-value will be the maximal area:k = A(125) = –(125)2 + 250(125) = –15 625 + 31 250 = 15 625The problem didn't ask me "what is the value of the variable w?", but "what are the dimensions?" I have w = 125. Then the length is L = 250 – w = 250 – 125 = 125.

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