Sunday Night Calculus Thread

[MrCodeDude]

A right triangle in the first quadrant has the coordinate axes as sides, the hypotenuse passes through the point (1,8). Find the vertices of the triangle such that the length of the hypotenuse is a minimum.

 

[Ender]

N/M.

 

[MrCodeDude]

I'm thinking.

I have to have the points

(0,0), (x,0) and (0,y)

But other than that, I don't know where to go.

Hypotenuse C = sqrt [ x^2 + y^2 ]

 

[MrCodeDude]

bump.

 

[MrCodeDude]

Originally posted by: MrCodeDude
bump.

 

[BigPoppa]

If all else fails, plug and chug. I can't remember how to do this problem using related rates. Pick a few different slope values, find intercepts. Might jog your brain into how to do it using calc/related rates.

 

[MrCodeDude]

I've plugged and chugged and gotten four wrong answers 🙁

 

[raptor13]

The hypotenuse will be a line segment with slope y/x where (x,0) and (0,y) are two of your triangle's vertices. You know (1,8) is incident on the hypotenuse. You've not got a slope and a point. Make a line equation.

Length of the slope is sqrt(x^2 + y^2), you're right. It's also the distance between (x,0) and (0,y). Use the distance formula.

You now have two equations and two unknowns. Differentiate, solve. Done.

 

[Howard]

You need to use related rates for this?

Well, I suppose you should, to find the minimum.