The outer dimensions of a closed rectangular cardboard box are 8 centimeters by 10 centimeters by 12 centimeters, and the six sides of the box are uniformly 12 centimeter thick. A closed canister in the shape of a right circular cylinder is to be placed inside the box so that it stands upright when the box rests on one of its sides. Of all such canisters that would fit, what is the outer radius, in centimeters, of the canister that occupies the maximum volume

Vmax = 192.33 cm³

Step-by-step explanation: An error in the problem statement. The sides of the box could not be 12 cm. We assume 1.5 cm

Inside dimensions of the box:

Outer dimensions :          12          10         8

 2 *  1.5   =  3                      3            3         3

Inside dimensions:            9            7         5

The volume of a right circular cylinder is:

V(c)  =  π*r²*h              r is the radius of the base and  h the height

By simple inspection is obvious that volume maximum will occur when r is maximum, and r is maximum, only when the base of the cylinder is in the rectangle 12*10. ( Inside  dim  9*7 ) In that case r  =  7/2   r = 3.5 cm

Then the height is 5 cm.

And the maximum volume of the cylinder is:

Vmax = 3.14* ( 3.5)²*5

Vmax = 192.33 cm³

sorry not sure

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2