Consider a rectangle that is inscribed with its base on the x-axis and its upper corners on the parabola y=C−x2, with C>0. What are the width and height that maximize the area of this rectangle? What is that maximal area?
Accepted Solution
A:
Answer:\frac{4(9c-4)}{27Step-by-step explanation:Assuming c positive, we find that this parabola is open down with vertex on the y axis above the origin.If a rectangle is formed, then it would have two vertices on either side of the y axis of the parabola with remaining two vertices on x axis.Due to symmetry let us take vertices on x axis as (a,0) and (-a,0)Corresponding vertices on the parabola would be[tex](b, c-a^2) and (b,c-a^2)[/tex]Now the rectangle has width = 2b andlength = [tex]c-a^2[/tex]So area of the rectangle = [tex]A(a) =2a(c-a^2)\\A(a) = 2ac-a^3[/tex]Use derivative test to find a which gives maximum area[tex]A'(a) = 2a-3a^2\\A"(a) = 2-6a[/tex]Equate I derivative to 0 to geta =0 or a = 2/3a cannot be 0A" is negative for a = 2/3So maximum when a =2/3and maximum area = [tex]2(\frac{2}{3} )(c-\frac{4}{9})\\ =\frac{4(9c-4)}{27}[/tex]