Maximize area of triangle

Question

I'm trying to maximize the area of a triangle with the following three sides.
The first side is the y=0, the second lies on the line y = 3x, and the third passes through the point (1,1).

I want to find the slope of the third line that maximizes the area of the triangle. I used the equation A = ($ \frac{1}{2} b\cdot h $) and solving for the height in terms of the base.

I then combined those equations to get an equation for the area. When I maximize this and test my answers they don't seem to be working. Thanks for the help.

Answer 1

Hint: Use the two equations $$y=3x$$ and $$y=m(x-1)+1$$to compute the third point of the triangle.

Answer 2

There is no maximum, however there is a minimum area.

Consider the third side having slope $-\varepsilon$ where $0<\varepsilon\ll 1$ (so almost parallel to the $x$-axis). Then it has base $1+\varepsilon^{-1}$ (the vertices $(0,0)$ and $(1+\varepsilon^{-1},0)$ on the $x$-axis) and height $\approx 1$, so having area approximately proportional to $\varepsilon^{-1}\to\infty$ as $\varepsilon\to 0^+$.

Indeed the graph of the area as a function of the inverse-slope of side through (1,1) (constraint $\mu=m^{-1}<\frac13$) is a branch of an oblique hyperbola, so giving a minimum (at $\mu=-\frac13$ as your $b=\frac43$ indicates).

Answer 3

You can make triangles with the given conditions and the area as large as you wish.

Thus there is no maximum area for such triangles.

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Answer 4

You have two sides given of your triangle, the lines, $y=0$ and $y=3x$.

Use the point-slope form of a line to express the line passing through $(1,1)$: $\frac{y-y_0}{x-x_0}=m$, where $y_0=1$ and $x_o$=1. In this case $m$ is the slope we want to find.

We want to find where the line through $(1,1)$ intersects the other line boundaries, so you need to solve $3x=m(x-1)+1$ and $0=m(x-1)+1$. Solve for $x$ and $y$. Take the $y$ for where the line through $(1,1)$ intersects as your height and the $x$ coordinate where the line through $(1,1)$ intersects the x-axis as your base. Insert into the Area formula you mention above.

You should end up solving a quadratic equation in $m$.