Can anybody help me with this optimization problem?

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 $y = \sqrt{4 - x}$ $y=sqrt(4-x)$

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

$A = x y$ $A=xy$
and
$A = x (\sqrt{4 - x})$ $A=x(sqrt(4-x))$

But I don't know how to continue

Thank you!

1 Answer

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Answer 1 · 2017-03-04T17:34:33

Dimensions of largest rectangle are $\frac{8}{3}$ $8/3$ and $\frac{2}{\sqrt{3}}$ $2/sqrt3$ and its area is $3.08$ $3.08$

Explanation:

By largest one means largest area.

As area is given by $A = x \sqrt{4 - x}$ $A=xsqrt(4-x)$

it will be maximized when $\frac{d A}{d x} = 0$ $(dA)/(dx)=0$ and $\frac{d^{2} A}{{d x}^{2}} < 0$ $(d^2A)/(dx^2)<0$

As $A = x \sqrt{4 - x}$ $A=xsqrt(4-x)$ , using product rule

$\frac{d A}{d x} = 1 \times \sqrt{4 - x} + x \times (\frac{1}{2} \times \frac{1}{\sqrt{4 - x}} \times (- 1))$ $(dA)/(dx)=1xxsqrt(4-x)+x xx(1/2xx1/sqrt(4-x)xx(-1))$

= $\sqrt{4 - x} - \frac{x}{2 \sqrt{4 - x}}$ $sqrt(4-x)-x/(2sqrt(4-x))$

and $\frac{d^{2} A}{{d x}^{2}} = - \frac{x}{2 \sqrt{4 - x}} - \frac{2 \sqrt{4 - x} - 2 x (\frac{- 1}{2 \sqrt{4 - x}})}{4 (4 - x)}$ $(d^2A)/(dx^2)=-x/(2sqrt(4-x))-(2sqrt(4-x)-2x((-1)/(2sqrt(4-x))))/(4(4-x))$

or $- \frac{x}{2 \sqrt{4 - x}} - \frac{\sqrt{4 - x} + \frac{x}{2 \sqrt{4 - x}}}{2 (4 - x)}$ $-x/(2sqrt(4-x))-(sqrt(4-x)+x/(2sqrt(4-x)))/(2(4-x))$

= $- \frac{x}{2 \sqrt{4 - x}} - \frac{2 (4 - x) + x}{4 \sqrt{4 - x} (4 - x)}$ $-x/(2sqrt(4-x))-(2(4-x)+x)/(4sqrt(4-x)(4-x))$

= $- \frac{x}{2 \sqrt{4 - x}} - \frac{8 - x}{4 {(4 - x)}^{\frac{3}{2}}}$ $-x/(2sqrt(4-x))-(8-x)/(4(4-x)^(3/2))$

and $\frac{d A}{d x} = 0$ $(dA)/(dx)=0$ , when $\sqrt{4 - x} = \frac{x}{2 \sqrt{4 - x}}$ $sqrt(4-x)=x/(2sqrt(4-x))$

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

Dimensions of largest rectangle are $\frac{8}{3}$ $8/3$ and $\sqrt{\frac{4}{3}}$ $sqrt(4/3)$

and its area is $\frac{8}{3} \sqrt{4 - \frac{8}{3}} = \frac{3}{8} \sqrt{\frac{29}{8}} ≅ 3.08$ $8/3sqrt(4-8/3)=3/8sqrt(29/8)~=3.08$

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

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Can anybody help me with this optimization problem?

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