Find the maximal area of a quadrilateral inside of a rectangle

Question

Problem:

In a rectangle $ABCD$ we know that $AB=3$ and $BC=1$. If $M$ is a point of side $AD$ and $N$ is a point of side $AB$, such that $AN=2AM$, what is the maximal possible area of the quadrilateral $MNCD$?

Solution:

$$\frac{7}{4}$$

How do we even approach this problem?

Answer 1

Think about the triangles $MAN$ and $NBC$, depicted above.

Triangle $MAN$ has area $x^2$ and triangle $NBC$ has area $\frac{3}{2} - x$.

The area of the quadrilateral $MNCD$ is the area of the rectangle $ABCD$ minus the areas of these triangles, thus $MNCD$ has area $\frac{3}{2} + x - x^2$.

This is now a calculus problem: maximize $\frac{3}{2} + x - x^2$ subject to $0 \leq x \leq 1$.

Answer 2

Let $AM=a$.

Thus, $AN=2a$, $NB=3-2a$ and the area it's $$3\cdot1-\frac{1}{2}a\cdot2a-\frac{1}{2}1\cdot(3-2a)=-a^2+a+\frac{3}{2}=$$ $$-a^2+a-\frac{1}{4}+\frac{3}{2}+\frac{1}{4}=-\left(a-\frac{1}{2}\right)^2+\frac{7}{4}\leq\frac{7}{4}.$$ The equality occurs for $a=\frac{1}{2},$ which says that we got a maximal value.