Quiz11Bsol - Solutions to Quiz 11B www.math.ufl.edu/˜harringt 1 Mrs Zerta wants to start a garden next to her house She only has 400 yards of

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Unformatted text preview: Solutions to Quiz 11B www.math.ufl.edu/˜harringt April 11, 2008 1. Mrs. Zerta wants to start a garden next to her house. She only has 400 yards of fencing. If fencing is NOT required along side of her house, what are the dimensions of the largest area that she can enclose? What is this area? Solution: We will let x denote the width and y denote the length. So we have this image: So we have that the perimeter is P = 2x + y = 400. Also recall that the area is: A = xy . Thus, we will solve for y using the perimter. So, y = 400 − 2x. Next, we will substitute y = 400 − 2x back in the formula for area. Hence, A = x(400 − 2x) = 400x − 2x2 . This is the equation we want to maximize. Also note that, x ≥ 0, and y = 400 − 2x ≥ 0. So, 0 ≤ x ≤ 200. Therefore, we want to find the absolute max for A = 400x − 2x2 on the interval on [0, 200]. To find the critcial number we set A = 0 so A (x) = 400 − 4x = 0 implies that x = 100. (Note: if we check the enpoints we just get A(0) = 0 and A(200) = 0.) So y = 400−2(100) = 200. Thus the dimensions are 100 yards by 200 yards. The area is A = 100 ∗ 200 = 20000 yrds2 . 2. Supose that g (x) = ex and g (0) = −1, then what is g ? So g (x) = ex + c. Since g (0) = −1 then e0 + c = 1 + c = −1 implies c = −2. Thus, g (x) = ex − 2 3. Suppose that f (x) = x2 − 5x and f (1) = 1, then what is f ? So f (x) = 1 x3 − 5 x2 + c. 3 2 Since f (1) = 1, then 1 − 5 + c = 1 implies c = 19 . Thus, 3 2 6 5 19 1 f (x) = x3 − x2 + 3 2 6 1 ...
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This note was uploaded on 07/10/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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