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I'm having trouble with this basic math question. Any hints?

Given a graph with lines $x=y$ and $x+3y=4$. Consider the triangle formed by these two lines and $y=0$.

When $y$ is increasing from $0$ to $1$, how is the length of $y$ (slice inside the triangle) decrease? As in what function does that describe? For example, the slice length is $4$ when $y=0$ and $0$ when $y=1$.

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Please check the second line. To be consistent with next paragaph, it should be "and $y=0$" not "and $x=0$". –  André Nicolas Dec 13 '11 at 2:27
    
True; changed it. Any ideas? –  user21128 Dec 13 '11 at 2:32
    
Yes, the slice length decreases linearly. Draw a picture. One line is $x=4-3y$, another is $x=y$. From the picture, slice is $(4-3y) - y$, that is, $4-4y$. –  André Nicolas Dec 13 '11 at 2:36
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A picture should accompany this verbal description.

The triangle we are discussing has its vertices (going counterclockwise) at $(0,0)$, $(4,0)$, and $(1,1)$.

Take a slice of this triangle, parallel to the $x$-axis, at height $y$. The rightmost end of the slice is on the line $x+3y=4$, and the leftmost end of the slice is on the line $x=y$.

The slice meets the line $x+3y=4$ at the point with $x$-coordinate $x_r=4-3y$. The slice meets $x=y$ at (where else) the point with $x$-coordinate $x_l=y$.

The length of the slice is $x_r-x_l$. This is $(4-3y)-y$, or more simply $4-4y$.
This function is linear in $y$, at least on the interval where it makes geometric sense, which is $0 \le y \le 1$.

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Makes sense. smacks head Then the marginal distribution of y will be 4-4y when 0<y<1. Right? –  user21128 Dec 13 '11 at 3:51
    
@user21128: You described the problem in terms of analytic geometry. I thought that perhaps it was preliminary to a volume calculation by the Method of Cylindrical Shells. Now that you mention marginal distribution, it looks as if the setting is probabilistic. As a guess you have a pair $(X,Y)$ of random variables, whose joint distribution is uniform on our triangle. If so, since the triangle has area $2$, the density function is $\frac{1}{2}$ on or inside the triangle, $0$ outside. For the marginal distribution, you need to multiply our $4-4y$ by $\frac{1}{2}$, getting $2-2y$. –  André Nicolas Dec 13 '11 at 4:11
    
Oh yes. I forgot to divide by area. Thank you very much sir! –  user21128 Dec 13 '11 at 7:06
    
@user21128: Happy to be of assistance. If you thought the answer was $4-4y$, integrating from $0$ to $1$ would give you $2$, which cannot be right, since the integral of a density function must be $1$. So a (partial) check is always available. –  André Nicolas Dec 13 '11 at 7:32
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