Find the dimensions of the rectangular box with the largest possible volume that you can make with 12 squared meters of cardboard. Assume it is closed on all sides.
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The Volume of a box with a square base x by xcm and height h cm is V=x2h
The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is A=x2+4xh
We need A as a function of x alone, so we'll use the fact that
V=x2h=32,000 cm^3
which gives us h=32,000x2, so the area becomes:
A=x2+4x(32,000x2)=x2+128,000x
We want to minimize A, so
A'=2x−128,000x2=0 when 2x3−128,000x2=0
Which occurs when x3−64,000=0 or x=40
The only critical number is x=40 cm.
The second derivative test verifies that A has a minimum at this critical number:
A''=2+256,000x3 which is positive at x=40.
The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is A=x2+4xh
We need A as a function of x alone, so we'll use the fact that
V=x2h=32,000 cm^3
which gives us h=32,000x2, so the area becomes:
A=x2+4x(32,000x2)=x2+128,000x
We want to minimize A, so
A'=2x−128,000x2=0 when 2x3−128,000x2=0
Which occurs when x3−64,000=0 or x=40
The only critical number is x=40 cm.
The second derivative test verifies that A has a minimum at this critical number:
A''=2+256,000x3 which is positive at x=40.
spark25:
hi
I have a doubt that what is the area of a trapezium
pls
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