Find the dimensions of the rectangle of maximum area bounded by the x-axis and y-axis and the graph of
A) length 1.5; width 1.25
B) length 2; width 1
C) length 0.5; width 1.75
D) length 1; width 1.5
E) none of the above

Question

Find the dimensions of the rectangle of maximum area bounded by the x-axis and y-axis and the graph of   ​
A) length 1.5; width 1.25 
B) length 2; width 1 
C) length 0.5; width 1.75 
D) length 1; width 1.5 
E) none of the above

Quizplus · Accepted Answer

The graph of the function is a parabola with vertex at (0,0) and opening downwards. The x-intercepts are (-2,0) and (1,0). The y-intercept is (0,4). Therefore, the rectangle of maximum area must have two points on the x-axis and two points on the y-axis, and the vertices must be on the x-intercepts and y-intercept. Let the length of the rectangle be 2x and the width be 2y. Then the area is A = 4xy. Since the rectangle is symmetric along the y-axis, we can focus on the first quadrant, where x and y are both positive. Let's first consider the case where one vertex of the rectangle is (1,0). Then the other two vertices are (-x,y) and (-x,-y) for some positive x and y. Since (-x,y) and (-x,-y) are on the parabola, we have 4y = x^2 - 2x and 4y = x^2 + 2x. Adding the two equations, we get x^2 = 8y. Substituting into A = 4xy, we get A = 32y^2/4 = 8x^2. Thus, we want to maximize x^2 subject to the constraints x > 0 and 4y = x^2 - 2x. Using Lagrange multipliers, we have f(x,y) = x^2, g(x,y) = x^2 - 2x - 4y, and L(x,y,lamda) = x^2 + lamda(x^2 - 2x - 4y). Taking partial derivatives, we get 2x + 2lamda(x - 1) = 0 and -4lamda = 0. Since lamda must be negative (otherwise we get a minimum), we have x = 1/2 and y = 7/8. Therefore, the dimensions of the rectangle are 2x = 1 and 2y = 7/4, so the answer is B.

Find the Dimensions of the Rectangle of Maximum Area Bounded