Calculus Optimization question

Question

 What is the largest area possible for a rectangle contained between the x-axis and the quadratic with y-intercept of (0,10) and x-intercepts of (-5,0) and (5,0)

 

Answer 1

Zeroes of the quadratic are -5 and 5, so:

y=a(x-5)(x+5)=ax²-25a. When x=0, y=10, so -25a=10, a=-⅖.

Therefore y=-⅖x²+10 is the equation of the quadratic (parabola).

Consider a rectangle ABCD, where A=(-x,0), B=(-x,y), C=(x,y), D=(x,0), then its dimensions are 2x by y, area A=2xy=2x(10-⅖x²)=20x-⅘x³.

dA/dx=20-12x²/5=0 for max/min. d²A/dx²=-24x/5, which is negative because x>0, so this is a maximum.

Therefore 12x²/5=20, ⅗x²=5, x²=25/3, x=5√3/3.

Area=20x-⅘x³=38.49units² approx.