*Optimization problem* [I don't know how to get to the answer] Problem: You have a cylindrical can with radius 4cm and height 10cm. Inside is a marble with a radius that has to be larger than 0 but less than 4 cm (even the largest marble will fit entirely). You're filling it with water and you stop when the height of water reaches the top of the marble. find the radius of the marble requiring the most water to cover it. Hint: Volume of water in a cylinder with radius R and height of water H is given by V=PiR^2^H and volume of a sphere with radius r is given by V=4/3Pi(r^3^). Consider: if the water is just covering the marble of radius r, what is the height of the water in the can?

Question

*Optimization problem* [I don't know how to get to the answer] Problem: You have a cylindrical can with radius 4cm and height 10cm. Inside is a marble with a radius that has to be larger than 0 but less than 4 cm (even the largest marble will fit entirely). You're filling it with water and you stop when the height of water reaches the top of the marble. find the radius of the marble requiring the most water to cover it. Hint: Volume of water in a cylinder with radius R and height of water H is given by V=PiR^2^H and volume of a sphere with radius r is given by V=4/3Pi(r^3^). Consider: if the  water is just covering the marble of radius r, what is the height of the water in the can?

Steve · Answer

the volume of water when the marble is just covered is v = pi * 4^2 * 2r - 4/3 pi r^3 = 32pi r - 4/3 pi r^3 dv/dr = 32pi - 4pi r^2 dv/dr = 0 when r = √8 since d2v/dr2 < 0, this is a max.

Ken · Answer

Thank you! Question: how did you reach d2v/dr2? (did you take a second derivative?)

Steve · Answer

that's what it is. It is used to determine whether an extremum is a max or min.

Ian · Answer

How did you get 2r for the height in the volume of water?

Steve · Answer

come on, guy! the diameter of the ball is the depth of the water!