**Exercise:** Find the dimensions of the rectangle of largest area that can be inscribed in an equilateral triangle of side L if one side of the rectangle lies on one side of the triangle.

Suppose the length of the side of the rectangle that lies on one side of the triangle x, and the length of another side y. We want to maximize the area of the rectangle A=xy. However, x+\frac{2}{\sqrt{3}}y=L. Thus A=(L-\frac{2}{\sqrt{3}}y)y. Take the derivative, we get A'=-\frac{2}{\sqrt{3}}y+(L-\frac{2}{\sqrt{3}}y)=L-\frac{4}{\sqrt{3}}y. Then if y=\frac{\sqrt{3}}{4}L and x=L-\frac{2}{\sqrt{3}}y=\frac{1}{2}L, then area reaches its maximum.

**Exercise:** A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at $10, the average attendance had been 27,000. When ticket prices were lowered to $8, the average attendance rose to 33,000.

(a) Find the demand function, assuming that it is linear.

(b) How should ticket prices be set to maximize revenue?

(a) The demand function d(p)=1000(57-3p), where p is the ticket price.

(b) The revenue R=p\times d(p)=1000p(57-3p). Similar to the previous problem, if we take the derivative, we can see that if fix the price at $9.5, the revenue will be maximized.

**Exercise:** Draw the graph of y=\frac{x^2-4}{x}.

Sketch can be found here.

**Exercise:** If 1200 cm^{2} of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Suppose the length of the sides of the square base is x cm and the height of the box is y cm. We want to maximize the volume of the box V=x^2y. However, x^2+4xy=1200. This gives us xy=300-x^2/4. Then V=x(xy)=x(300-x^2/4). Take the derivative, we get V'=300-x^2/4+x(-x/2)=300-\frac{3}{4}x^2. If x=20, it maximizes the volume of the box, which is 4000 cm^{2}.