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Unformatted text preview: Mathematics 250 Fall 2008 Selected Solutions Assignment # 7 MVM 5.2.10 : Find the the isosceles trapezoid with maximum area, given the total length of the two sides and the bottom. Response: There are three variables involved in this maximization problem: the length of the base, the common length of the two sides, and the height. The height is determined by the angle between the sides and the base, so one could also use the length of the base, the length of (each of) the two sides, and the angle between the base and the sides. This is suggested in the problem statement. Let b denote the length of the base, and let a denote the length of the two sides. Then a and b satisfy the constraint that 2 a + b = w , where w is the total length of the three sides. (As stated, the problem specifies that w = 12 inches.) Following the problem suggestion, let θ be the (exterior) angle between the side and the base. Then the side of the trapezoid is the diagonal of a rectangle with horizontal base equal to a cos θ and height equal to a sin θ . Thus, the length of the top of the trapezoid is b + 2 a cos θ and the height of the trapezoid is a sin θ . Therefore, the area of the trapezoid is A = 1 2 ( b + b + 2 a cos θ ) a sin θ = a ( b + a cos θ ) sin θ. One way to approach this problem is to use the fact that the variable θ is not involved in the constraint, so, for any given b and a , one can find the value of θ that maximizes the area of the trapezoid. This maximum area will be a function of a and b . We can think about maximizing this function while satisfying the constraint b + 2 a = w . Therefore, we compute ∂A ∂θ = a ( b + a cos θ ) cos θ- a 2 sin 2 θ = ab cos θ + a 2 (cos 2 θ- sin 2 θ ) = 2 a 2 cos 2 θ + ab cos θ- a 2 . This is a quadratic expression in cos θ . Setting it equal to zero and solving gives cos θ = √ 8 a 2 + b 2- b 4 a . From this we can derive an expression for sin θ . Then we can plug both these expressions into the formula for A , obtaining an expression for A involving only a and b , We can also replace b with w- 2 a , getting an expression for A involving only a , which we can then attempt to maximize. However, the expression for A that we get will be messy. In some sense, we are doing too much work when we seek to adjust θ to maximize the volume for bad choices of a and b . The value of θ that works for the correct choice of a and b may be nice, whereas we can get many values of θ if we choose a and b at random. Therefore, we try the method of Lagrange multipliers. We want to maximize the function A of a , b , and θ , subject to the constraint that 2 a + b = w . We compute DA = [ b sin θ + 2 a sin θ cos θ a sin θ 2 a 2 cos 2 θ + ab cos θ- a 2 ] ....
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This note was uploaded on 09/24/2009 for the course MATH 250 taught by Professor Rogerhowe during the Fall '06 term at Yale.
- Fall '06