# Maximizing/Minimizing triangle area

I think I set it up right and I know area of a triangle is 1/2(BH). So I get 1/2(xy) and then try to maximize/minimize X? What do I plug in for Y?

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Please use parentheses when using the slash for division. Sometimes when we see $1/2(BH)$ it means $\frac 1{2BH}$, sometimes (as here) it means $\frac {BH}2$. Either $BH/2$ or $(1/2)BH$ or \frac {BH}2 between dollar signs to get the stacked fraction (preferred). Some hints are here –  Ross Millikan Nov 23 '13 at 4:18

## 2 Answers

Since the point is on the curve $y=\exp(\frac {-x}3)$ that is what you should use for $y$. Then you can maximize over $x$.

Added: The area of the triangle is $A=\frac 12x\exp(\frac {-x}3)$ Then $\frac{dA}{dx}=\frac 12\left( \exp(\frac{-x}3)-\frac x3\exp(\frac{-x}3)\right)$ Does that help?

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The coordinates of the vertices of the right triangle are (0,0) , (x,0), (x,y) but they tell you that the third point is on the curve y=Exp[-x/3]. So, the coordinates of the vertices are (0,0) , (x,0), (x,Exp[-x/3]). So the area is given by S = x Exp[-x/3] / 2. Since you want to find extremum values of theis area, its derivative (with respect to x, the only unknown) must be zero. Are you able to continue with this ?

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Okay so I found the derivative and set it = to 0 and got x=6, which is out of the domain? –  user104827 Nov 23 '13 at 4:47
Yeah nevermind I made some algebra mistakes, but I'm still left with the wrong answer. My derivative is ((e^-x/3)/2)-((xe^-x/3)/6) –  user104827 Nov 23 '13 at 4:53
Factor Exp[-x/3] in order to simplify your derivative (which is correct). You should not get x=6 as a solution but x=? –  Claude Leibovici Nov 23 '13 at 4:58
@RossMillikan. Thanks for reporting the typo –  Claude Leibovici Nov 23 '13 at 5:15
@user104827. Do not forget that you also must find the smallest area. –  Claude Leibovici Nov 23 '13 at 6:01