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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

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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

#1
+91462
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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 Sort: ### 1+0 Answers #1 +91462 +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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