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]}