*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?
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.
Thank you! Question: how did you reach d2v/dr2? (did you take a second derivative?)
that's what it is. It is used to determine whether an extremum is a max or min.
How did you get 2r for the height in the volume of water?
come on, guy! the diameter of the ball is the depth of the water!
To solve this optimization problem, we need to find the radius of the marble that requires the most water to cover it.
Let's consider the scenario where the water is just covering the marble of radius r. In this case, we need to find the height of the water in the can.
The height of the water in the can is the total height of the can minus the height of the marble. Since the height of the can is given as 10 cm and the height of the marble is an unknown value, we can represent the height of the water as H = 10 - h, where h is the height of the marble.
Now, we can calculate the volume of the water in terms of the marble's radius r using the volume formula for a cylinder: V = πr^2H.
Substituting the value of H, we have V = πr^2(10 - h).
We also know that the volume of a sphere with radius r is given by V = (4/3)πr^3.
Since the volume of the water in the can must be equal to the volume of the marble, we can equate the two equations:
πr^2(10 - h) = (4/3)πr^3.
Simplifying this equation, we get:
10 - h = (4/3)r.
Rearranging terms, we have:
h = 10 - (4/3)r.
So the height of the water in terms of the marble's radius r is h = 10 - (4/3)r.
Now, the volume of the water in the can is given by V = πr^2h.
Substituting the expression for h, we have V = πr^2(10 - (4/3)r).
To find the radius of the marble that requires the most water to cover it, we need to maximize this volume function.
To do so, we can differentiate V with respect to r, set the derivative equal to zero, and solve for r:
dV/dr = π(20r - (8/3)r^2) = 0.
Simplifying the equation, we get:
20r - (8/3)r^2 = 0.
Multiplying through by 3 to eliminate fractions, we have:
60r - 8r^2 = 0.
We can further simplify by factoring out r:
r(60 - 8r) = 0.
Setting each factor equal to zero, we have two possible values for r:
r = 0 (which is not a valid solution since the radius of the marble cannot be zero), and
60 - 8r = 0.
Solving for r in the second equation, we get:
60 - 8r = 0
8r = 60
r = 7.5.
So, the radius of the marble requiring the most water to cover it is 7.5 cm.
To summarize the steps to get the answer:
1. Express the height of the water in terms of the marble's radius.
2. Set the volume of water equal to the volume of the marble and simplify the equation.
3. Differentiate the volume function with respect to the marble's radius.
4. Set the derivative equal to zero and solve for the radius.
5. Check that the solutions are valid (in this case, discard r = 0).
6. The valid solution is the radius of the marble requiring the most water to cover it, which in this case is 7.5 cm.