surely you can graph a simple parabola.
Let the rectangle have height 2y and width a-x
The area is then
A = 2xy = 2(a - y^2/4p)y = 2ay - y^3/2p
dA/dy = 2a - 3y^2/2p
dA/dy = 0 when y = 2√(ap/3)
now just figure A=2xy there.
Find the dimensions of the rectangle of maximum area A that can be
inscribed in the portion of the parabola y^2=4px intercepted by the line x=a. heres another problem pls help Damon!!! Graph it if possible
2 answers
can you pls show me the graph and why the width = a-x? pls reply