# Maximum area of a bounded rectangle

Without calculus, I am trying to find the maximum area of a rectangle that is bounded by the $x$ and $y$ axis and bounded by the line $y=-2x+1$. It is also parallel to both axis.

I would post an attempt but I am lost on how to even get this started...

Updated attempt

Area = $xy$ $$= x(-2x+1)$$ $$= -2x^2+ 2x$$

This parabola opens down so it has a maximum at its vertex which is $(1/4, 3/8)$

## 3 Answers

Hint: let $a,b$ be the coordinates of the rectangle vertex opposite to the origin, then $a, b \ge 0$ and $b=-2a+1$. The area is:

$$a \cdot b = a (1-2a)=-2a^2 + a = -2\left(a-\frac{1}{4}\right)^2 + \frac{1}{8} \le \frac{1}{8}$$

• I followed up to where you got $-2a^2-a$ What did you do from there? – combo student Jan 8 '18 at 2:38
• @combostudent Complete the square: $-2a^2+a=-2\left(a^2-2 \cdot \frac{1}{4} \cdot a + \color{red}{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2}\right)$ $=-2\left(\left(a-\frac{1}{4}\right)^2-\frac{1}{16}\right)$ $=-2\left(a-\frac{1}{4}\right)^2 + \frac{1}{8}\,$. – dxiv Jan 8 '18 at 2:44
• Got it and why is that less than or equal to $1/8$? – combo student Jan 8 '18 at 2:45
• @combostudent Because the square of a real number is always non-negative, so $-2\left(a-\frac{1}{4}\right)^2 \le 0\,$ (with equality iff $\,a = \frac{1}{4}\,$). Now add $\,\frac{1}{8}\,$ to both sides of the previous inequality. – dxiv Jan 8 '18 at 2:47
• Can I get the vertex of this parabola to find my maximum x value then Plug that in the equation of the line to get the y value ? – combo student Jan 8 '18 at 3:37

The area it's $$x(1-2x)$$ and since $0<x<\frac{1}{2}$, we can use AM-GM: $$x(1-2x)=\frac{1}{2}\cdot2x(1-2x)\leq\frac{1}{2}\left(\frac{2x+1-2x}{2}\right)^2=\frac{1}{8}.$$ The equality occurs for $2x=1-2x,$ which says that $\frac{1}{8}$ is a maximal value.

Hints: Put one corner at the origin, one on the line $y=-2x+1$. Call that second corner $(a,b)$. Express the area in terms of $a$ and $b$, and use the line equation to write $b$ in terms of $a$.

Thus, write the area as a function of $a$ only. Maximize that function with whatever tools you've got (knowledge of quadratic functions is sufficient).

• I know that the vertex of a parabola is either the max or min. Does that help? – combo student Jan 8 '18 at 2:24
• Sure! Use all your tools for graphing the quadratic function that you get to see what value of $a$ will give you the maximum area. Cheers! – Matthew Conroy Jan 8 '18 at 3:59