rectangles beneath a line a. A rectangle is constructed with one side on the po ? sitive? -axis one side on the positive ? y-axis, and the vertex opposite the origin on the line ? y? =10 ? 2?x. What dimensions maximize the area of the rectangle? What is the maximum area? b. Is it possible to construct a rectangle with a greater area than that found in part (a) by placing one side of the rectangl?e on the li?ne ?y =10 ? 2x ? , and the vertices not on that line on? the p? ositive? and y? -axes? Find the dimensions of the rectangle of maximum area that can be constructed in this way.
Solution 24E \Step 1: (a)Let us consider the side on the x-axis extends to the point (x,0) and the side on the y-axis to (0,x).Line equation which has vertex on it is given by y=10-2x Consider the length of the rectangle be x and the width of rectangle be y then area of rectangle,A=xy The objective function is area of rectangle A=xy Step 2: We convert A in terms of single variable x A=x(10 2x) =10x 2x 2 ;0 x 5 Which is the function in single variable x With A(0)=10(0)-2(0)=0 2 And A(5)=10(5)-2(5) =50-50 =0 Step 3: Now in order to maximize A we need to find critical point by evaluating A’=0 A’=0 dA = 0 dx d (10x 2x ) = 0 dx (10 4x) = 0 x = 14 = 2 Hence x= 5 gives maximum area 2