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Unformatted text preview: SOLUTIONS CHAPTER 15.1 MATH132 WI01 4. Call the length of the big rectangle L and the width W . The total length of fence needed, let’s call it Y , is the rectangle’s perimeter PLUS the lengths of the 3 inner fences → (2 L + 2 W ) + 3 W = 2 L + 5 W . What we need now is a relation between L and W : since the area encompased by the fence is 1000, and can be obtained using formula for area of a rectangle, namely area = L * W → L * W = 1000 → W = 1000 L . Now we can plug this formula for W in the Y formula → Y = 2 * L + 5 * W = 2 L + 5 * 1000 L = 2 L + 5000 L = 2 L 2 L + 5000 L = 2 L 2 +5000 L . We’re looking for least length hence we need the minimum for the Y . Compute derivative of Y with respect to L (that is, L is the variable): Y = 2 * 2 L * L (2 L 2 + 5000) L 2 = 4 L 2 2 L 2 5000 L 2 = 2 L 2 5000 L 2 . Y = 2( L 2 2500) L 2 = 2( L 50)( L + 50) L 2 . For the minimum we need critical numbers, that is zeroes for Y , so L = ± 50, and also Y =DNE, hence L = 0. But we have some practical reasons to ignore= 0....
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 Spring '08
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 Math, Calculus, Critical Point, Height, Elementary algebra, critical numbers, MATH132 WI01

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