Question

Can anybody help me with this optimization problem?

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!

Answer 1

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

Explanation:

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]}