# Recitation 17

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 cm2 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 cm2.