A company manufactures cans of cheese designed a can that has a
radius of 2 inches and a volume of 150 cubic inches. For a new can
design, the company plans to keep the height the same and triple the
radius (it's a really wide can of cheese). What will be the effect on the
can's volume?

Correct option: second one

(Volume will be 9 times higher)

Step-by-step explanation:

The volume of a cylindrical can is given by the formula:

V = pi * r^2 * h

Where r is the base radius and h is the height.

If the new radius will be the triple of the old radius and the height will be the same, we have:

r' = 3r

h' = h

So, we can find the new volume V':

V' = pi * r'^2 * h'

V' = pi * (3r)^2 * h

V' = pi * 9r^2 * h = 9 * (pi * r^2 * h) = 9V

So we have that the new volume will be 9 times higher than the old volume.

Correct option: second one.

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.