# Tutor profile: Kyle L.

## Questions

### Subject: Geometry

You have a hollow cone with slant height 58 in. and base radius 40 in. You want to fit a cylinder of radius 20 in. entirely within the hollow cone. What is the maximum volume of the cylinder?

The maximum volume is achieved with the maximum height, or when the top of the cylinder meets some part of the cone. Drawing the cross section of when you cut the cone straight down through its vertex, you have a rectangle within a triangle. In other words, you have triangle ABC where A is the vertex of the cone, AB = AC = 58, and BC = 80. Let O be the midpoint of BC. Rectangle WXYZ is inside ABC such that W is on AB, X is on AC, and Y and Z are on BC. Essentially, we are interested in WZ which is the height of the cone. By pyhagorean theorem, we know that AO = 42. We also notice that triangle ABO is similar to triangle WBZ because of congruent angles. Since BO = 40 and ZO = 20, BZ = 20. Setting up a proportion for side lengths, we have: $$\frac{AO}{BO} = \frac{WZ}{BZ}$$ $$ \frac{42}{40} = \frac{WZ}{20}$$ WZ = 21 Now to solve for the volume of the cylinder. $$\pi r^2h = \pi 20^2 * 21 = 8400 \pi$$

### Subject: Calculus

What is the between the x-axis and the curve $$x^3 - \frac{3}{2}x^2 - 18x + 100$$ from the curve's local maximum to the curve's local minimum.

The find the area between the x-axis and the curve, we will need to integrate. To find the bounds of integration, we will first need to find the local min and max using derivatives. $$\frac{d}{dx}(x^3 - \frac{3}{2}x^2 - 18x + 100) = 3x^2 - 3x - 18$$ $$ 3x^2 - 3x - 18 = 3(x+2)(x-3)$$ This is 0 when $$x=-2$$ and $$x = 3$$. Therefore our bounds of integration are -2 and 3. $$\int_{-2}^{3} x^3 - \frac{3}{2}x^2 - 18x + 100 dx = 453.75$$

### Subject: Algebra

You exist in a magical universe standing on the point (0,0). The ground is represented by the horizontal line y = 0. In this magical universe, there exist a magical balls that may fly above and below ground, ignoring the laws of physics. The magical ball follows the trajectory: $$x^6 - 64$$. Find the sum of the distances between you and each point the magical ball contacts the ground.

Since the ground is defined as $$y=0$$, you are interested in the roots of $$x^6 - 64$$. The quick way to solve this problem is to realize that $$x^6 - 64$$ is an even function meaning it is symmetrical with respect to the y-axis. Since you stand on (0,0), you are on the y-axis. Therefore, the sum of the distances between you and all the roots must be 0 The long way is to factor $$x^6 - 64$$ $$x^6 - 64 = (x^3 - 32)(x^3 + 32)$$ $$(x^3 - 8)(x^3 + 8) = (x-2)(x^2+2x+2)(x+2)(x^2-2x+2)$$ This is 0 when $$x=2$$ and when $$x=-2$$. Therefore the answer is 0.

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