Maximum dimensions-very interesting

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The discussion centers on designing a container box from a square cardboard sheet with an area of 1600 cm², specifically focusing on maximizing the volume of the box. The calculations show that the side length of the cardboard is 40 cm, and by cutting out squares from the corners, the volume formula is derived as V = (40-2a)².a. To find the maximum volume, the derivative dV/da is set to zero, leading to critical points at a = 20 and a = 20/3, with the latter providing the maximum volume. The optimal dimensions for the box are determined to be a square base of side 80/3 cm and a height of 20/3 cm. The discussion emphasizes the importance of confirming that a = 20/3 yields a maximum volume for completeness.
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maximum dimensions-very interesting!

hey guys...i have another question...i posted my answer right below it...does it look right guys?! I hope so! Thanks in advance

You are to design a container box by cutting out the four corners of a square cardboard sheet that is 1600cm^2 in area. The box must have a square base and an open top. Determine the dimensions of the box that give maximum value.

The side of the square cardboard is 40 cm. Let side of corner square is a cm. The volume of the box V = (40-2a)2.a cm3 , For max volume we have dV/da = 0 implies (20-3a)(20-a) = 0, i.e., a = 20, a = 20/3. But when a = 20, V = 0. So to maximize the volume a = 20/3. So the base of is a square of side 80/3 cm and height = 20/3 cm.

Sorry not good with latex...
 
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Yes, that looks good. It's intuitively clear that a=20/3 should be a maximum (there must be 'some' maximum between a=0 and a=20), but for completeness you might want to show it is a maximum indeed (depends on your teacher).
 
thanks for your help!
 
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