Exam3_2 6 - -2 x (12 x ) ( x 2 + 1) 2 = 12-12 x 2 ( x 2 +...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 6 is formed with adjacent sides on the coordi- nate axes and one corner on the graph of y = 12 x 2 + 1 Find the maximum possible area of this rect- angle. 1. max area = 5 sq. units. 2. max area = 4 sq. units. 3. max area = 8 sq. units. 4. max area = 6 sq. units. correct 5. max area = 7 sq. units. Explanation: The rectangle has side lengths x and y , so its area is given by A = xy . On the other hand, one corner lies on the graph of y = 12 x 2 + 1 so this area can expressed solely as a function by A ( x ) = x 12 x 2 + 1 · = 12 x x 2 + 1 . Di±erentiating with respect to x we see that A 0 ( x ) = 12( x 2 + 1)
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Unformatted text preview: -2 x (12 x ) ( x 2 + 1) 2 = 12-12 x 2 ( x 2 + 1) 2 . At a maximum A ( x ) = 0, i.e. , 12-12 x 2 ( x 2 + 1) 2 = 0 , i.e. , A ( x ) = 0 when x = ± 1. On practical grounds the positive solution corresponds to maximum area, while the negative solution has no meaning for this problem. Consequently, the maximum area = A (1) = 6 sq. units. . keywords: optimization, maximize, rectangle, area, word problem, 012 (part 1 of 1) 10 points If the graph of f is which one of the following contains only graphs of anti-derivatives of f ? 1. 2....
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.

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