An inbox tray has 3 walls and an open side on one of the longer sides.

Determine the maximum area of the tray if all three walls total to a length of 812 mm. Show your formula.

Guest Dec 3, 2014

#1**+5 **

Let the 2 short sides be x and the long side be y

$$\\2x+y=812\\

y=812-2x\\\\

Area(A)=x*y\\

A=x(812-2x)\\

A=812x-2x^2\\\\

\frac{dA}{dx}=812-4x\\\\

\frac{d^2A}{dx^2}=-4 \quad \cap\\

$This means that any stationary point will be a maximum$\\\\

$find stat point $\frac{dA}{dx}=0\\\\

0=812-4x\\

4x=812\\

x=203\\

$The tray will be of maximum are if the sides are 20.3cm by 40.6cm$$$

Melody
Dec 3, 2014

#1**+5 **

Best Answer

Let the 2 short sides be x and the long side be y

$$\\2x+y=812\\

y=812-2x\\\\

Area(A)=x*y\\

A=x(812-2x)\\

A=812x-2x^2\\\\

\frac{dA}{dx}=812-4x\\\\

\frac{d^2A}{dx^2}=-4 \quad \cap\\

$This means that any stationary point will be a maximum$\\\\

$find stat point $\frac{dA}{dx}=0\\\\

0=812-4x\\

4x=812\\

x=203\\

$The tray will be of maximum are if the sides are 20.3cm by 40.6cm$$$

Melody
Dec 3, 2014