A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y equals = 100 100 minus −x squared 2. What are the dimensions of the rectangle with the maximum area? What is that area?
Accepted Solution
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Answer:Step-by-step explanation:Given that a rectangle is constructed with its base on the x axis and two of its vertices on the parabola[tex]y=100-x^2[/tex]This parabola has vertex at (0,100) and symmetrical about y axis.Any general point above x axis can be written as (a,b) (-a,b) since symmetrical about yaxis. Hence coordinates of any rectangle are [tex](a,0) (-a,0), (a, 100-a^2), (-a, 100-a^2)[/tex]Length of rectangle = 2a and width = [tex]100-a^2[/tex]Area of rectangle = lw = [tex]2a(100-a^2)=200a-400a^3[/tex]To find max area, use derivative test.[tex]A' = 200-800a^2\\A"=-1600a<0[/tex]Hence maxima when first derivative =0i.e. when a =2Thus we find dimensions of the rectangle are l =4 and w = 96Maximum area = [tex]4(96) = 384[/tex]