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Hi so I am interested in finding the area of a graph given by this: $$\{(x,y)| 0 \leq x + y \leq \frac{1}{3}, \quad x\geq 0, \quad y\geq 0\},$$ graph between $0$ and $\frac{1}{3}$ for $x+y$, I've done a lot of calculus and integrations, but don't know where to start this this one.

Any help would be appreciated.

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Draw the graph of the region $\{(x,y)| 0 \leq x + y \leq 1/3\}$.

Note that it would be a region between the two straight lines $x+y=0$ and $x+y=1/3$ and if you intersect it with the first quadrant( As $x,y>0$), you will get a triangle, whose area is $1/2\times 1/3\times 1/3=1/18$.

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Hint. If you would like to solve the problem by using integration, consider the function $y=f(x)=1/3-x$. Then the area you are that looking for is given by $$A=\int_{0}^a f(x)\,dx$$ where $a$ is such that $y=f(x)\geq 0$ in $[0,a]$. What is $a$? What is $A$?

P.S. The region given by the inequalities is a right triangle (try to draw it), therefore you can compute the area also in another way (much simpler).

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  • $\begingroup$ Ah i see, this will be useful as I have other problems that get a little bit more complicated then just from 0 to 1. Thank you. $\endgroup$ – ZarifS May 22 '17 at 5:30
  • $\begingroup$ Actually, in this case the integration is from $0$ to $1/3$. $\endgroup$ – Robert Z May 22 '17 at 5:40
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Area of {$(x,y)$: $0 \le x +y \le 1/3; x,y \ge 0$}.

Consider the family of straight lines given by $x +y = c$ in the first quadrant.

$0 \le x +y = c \le1/3$.

So the $y-$intercept $c$ of the family of lines $y = - x +c$ , with slope $m = -1$, is between $0$ and $1/3$.

The maximal area of the triangle cut out by the positive $x,y$ - axes and the family of lines $y = - x + c$ , where $0 \le c \le 1/3 $ is

$1/2×1/3 ×1/3 =1/18$, where

$y$ - intercept = $1/3$, and $x$ -intercept = $1/3$

with $y = -x + 1/3$.

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