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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

Best Answer 

 #1
avatar+92174 
+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
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1+0 Answers

 #1
avatar+92174 
+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

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