# Tutor profile: Daniel S.

## Questions

### Subject: Geometry

A cylindrical can of soup has a radius of 2 inches and a height of 4 inches. What is the volume of the can of soup in square inches?

The formula for the volume of a cylinder is $$V=\pi r^2h$$. Since the radius is $$r=2 in$$ and the height is $$h=4 in$$, we can substitute these values into the first equation. This gives us $$V=\pi r^2h=\pi (2 in)^2 (4 in)=16 \pi in^3$$ as our Volume.

### Subject: Pre-Algebra

Simplify the inequality $$2x-3>7$$.

The inequality does not change if a number is added or subtracted from both sides. We can use this rule to add 3 to both sides of $$2x-3>7$$, giving us $$2x>10$$. The inequality also does not change if both sides are divided by a positive number. This means we can divide both sides of $$2x>10$$ by 2, giving us $$x>5$$, a much simpler answer.

### Subject: Algebra

A shepherd is building a fence to enclose his sheep inside a rectangular plot of land on the bank of a river, where the river forms one of the sides of the enclosure. The shepherd has 120 yards of material and wishes to maximize the area of the enclosure. Which values for the length and width, where the length is greater than the width, form an enclosure with the greatest possible area?

We create two variables for the length and width of the enclosure. We will define x as our width and y as our length. We will define our area as A. Since the river forms one of the sides of the rectangular piece of land, we know that $$2x+y=120$$. We are trying to maximize the value of A, where $$A=xy$$. We solve the first for y, giving us a solution of $$y=120-2x$$. We substitute this into the second equation, giving us Area $$A(x)=(120-2x)x=(-2)x^2+120x$$, as it is clear that the area can be expressed as a function of x. Since this equation represents a downward-facing parabola, we must find the vertex to determine which value of W gives us the greatest area. We use the formula $$x=\tfrac{-b} {2a}$$ to find our vertex, where a and b are coefficients of the quadratic equation $$A(x)=(-2)x^2+120x$$. This gives us $$x=-(120)/2(-2)=30$$. As $$x=30$$, we can substitute this value into the equation $$y=120-2x$$ to get $$y=120-2(30)=60$$. Since $$x=30$$ and $$y=60$$, we know that the dimensions that enclose the greatest area are: Length: 60 yards Width: 30 yards