Can anybody help me with this optimization problem?

Question

A rectangle has one vertex at the origin, one of the x-axis, one on the y-axis, and one on the graph of #y=sqrt(4-x)#

What is the largest the rectangle can have, and what are its dimensions?
This is everything I've figured out so far. I'm guessing that

#A=xy#
and
#A=x(sqrt(4-x))#

But I don't know how to continue

Thank you!

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Answer 1 · 2017-03-04T17:34:33

Dimensions of largest rectangle are #8/3# and #2/sqrt3# and its area is #3.08#

By largest one means largest area.

As area is given by #A=xsqrt(4-x)#

it will be maximized when #(dA)/(dx)=0# and #(d^2A)/(dx^2)<0#

As #A=xsqrt(4-x)#, using product rule

#(dA)/(dx)=1xxsqrt(4-x)+x xx(1/2xx1/sqrt(4-x)xx(-1))#

= #sqrt(4-x)-x/(2sqrt(4-x))#

and #(d^2A)/(dx^2)=-x/(2sqrt(4-x))-(2sqrt(4-x)-2x((-1)/(2sqrt(4-x))))/(4(4-x))#

or #-x/(2sqrt(4-x))-(sqrt(4-x)+x/(2sqrt(4-x)))/(2(4-x))#

= #-x/(2sqrt(4-x))-(2(4-x)+x)/(4sqrt(4-x)(4-x))#

= #-x/(2sqrt(4-x))-(8-x)/(4(4-x)^(3/2))#

and #(dA)/(dx)=0#, when #sqrt(4-x)=x/(2sqrt(4-x))#

or #2(4-x)=x# i.e. #8-2x=x# i.e. #x=8/3# and one can check that at #x=8/3#, #(d^2A)/(dx^2)<0#

Dimensions of largest rectangle are #8/3# and #sqrt(4/3)#

and its area is #8/3sqrt(4-8/3)=3/8sqrt(29/8)~=3.08#

Below is graph of #xsqrt(4-x)#
graph{xsqrt(4-x) [-3.063, 6.937, -1.12, 3.88]}