Which such triangle has the smallest area?

[Sheneron]

Homework Statement

A triangle has legs on the positive x and y axes, and its hypotenuse passes through the point (2,1). Which such triangle has the smallest area.

The Attempt at a Solution

I know the area of a triangle is 1/2 b*h. And I know that is the area that I am trying to minimize. So I could write the equation as A = 1/2 x*y. And to minimize that I will have to take the derivative and set it equal to 0. In order to do that I must rewrite the equation in terms of x's or y's. This is where I am stuck.

 

[Mark44]

You need to use the fact that the hypotenuse goes through (2, 1). The triangle's vertices are at (0, y), (x, 0) and the origin. Write the equation of the line that goes through the two intercepts and (2, 1). That should give you another relation between x and y so that you can get area as a function of one variable.

 

[Sheneron]

Would it be

$$1 = \frac{-2y}{x} + y$$

 

[Sheneron]

I think I have this so if someone would be so kind as to check it. And also a little edit, the point is (5,3).

$$A = \frac{1}{2}xy$$

From the given that the point is (5,3)
$$y = mx + b$$
$$3 = \frac{-y}{x}*5 + y$$
$$3 = y(\frac{-5}{x} + 1)$$

$$y = \frac{3}{\frac{-5}{x} + 1}$$

Now plugging that back into the area equation we have:

$$A(x) = \frac{1}{2}x(\frac{3}{\frac{-5}{x} + 1})$$

$$A(x) = \frac{3x}{\frac{-10}{x}+2}$$

So take the derivative of that and set it equal to 0 to find a minimum. That gets trick though so I graphed it and found the minimum to be at x = 10.

Plugging that back into the y equation gave an answer of y = 6.

So the base would be 10 and the height would be 6.