Question 114296
{{{drawing(600,400,-1,12,-1,8,
line(0,0,0,5),
line(0,0,10,0),
line(10,0,10,5),
locate(10.5,2.5, x),
locate(4.9,-.5,20-2x))}}}


The cross-sectional area is given by {{{A[cs]=f(x)=x(20-2x)=-2x^2+20x}}}


A critical point for f(x) is the value of x that satisfies {{{df(x)/dx=0}}}


{{{df(x)/dx=-4x+20}}}
{{{-4x+20=0}}}
{{{x=5}}}


Hence x = 5 is a critical point.  A critical point is a maximum if {{{d^2f(x)/dx^2 <0}}}


{{{d^2f(x)/dx^2 =-4<0}}}, therefore the critical point at x = 5 is a maximum.


Therefore, the depth of the gutter that will maximize its cross-sectional area and allow the greatest amount of water to flow is 5 inches.


The maximum cross-sectional area is the value of the area function at 5.


{{{A[max]=f(5)=-2(5)^2+20(5)=-50+100=50}}}


Therefore the maximum cross-sectional area is 50 square inches.


Hope that helps.
John