2
$\begingroup$

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 $(1/4, 3/8)$

$\endgroup$
1
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ I followed up to where you got $-2a^2-a$ What did you do from there? $\endgroup$ – combo student Jan 8 '18 at 2:38
  • $\begingroup$ @combostudent Complete the square: $-2a^2+a=-2\left(a^2-2 \cdot \frac{1}{4} \cdot a + \color{red}{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2}\right)$ $=-2\left(\left(a-\frac{1}{4}\right)^2-\frac{1}{16}\right)$ $=-2\left(a-\frac{1}{4}\right)^2 + \frac{1}{8}\,$. $\endgroup$ – dxiv Jan 8 '18 at 2:44
  • $\begingroup$ Got it and why is that less than or equal to $1/8$? $\endgroup$ – combo student Jan 8 '18 at 2:45
  • $\begingroup$ @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. $\endgroup$ – dxiv Jan 8 '18 at 2:47
  • $\begingroup$ 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 ? $\endgroup$ – combo student Jan 8 '18 at 3:37
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ I know that the vertex of a parabola is either the max or min. Does that help? $\endgroup$ – combo student Jan 8 '18 at 2:24
  • $\begingroup$ 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! $\endgroup$ – Matthew Conroy Jan 8 '18 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.