Question

Calculus Minimum and Maximum

Find the minimum and maximum values of the expression where x and y are lengths in the figure below and 0 ≤ x ≤ 2.

x+2y

minimum: ?
maximum: ?

My effort:
I create a main triangle and formed a (2-x) by (1) triangle, with 'y' as the hypotenuse. I solved for 'y' through the pythagorean theorem.
y^2 = (2-x)^2+1, y=sqrt((2-x)^2+1).

I substituted that into the equation 'x+2y' and it became 'x+2(sqrt......)'.

They're asking for minimum and maximum. I found the derivative, and got:
1- [ (4-x) / (sqrt((2-x)^2+1) ]

Set the derivative equal to zero, so

Answer 1

x+2y = what?

Answer 2

I tried the minimum to be:
0, 1, 9/8, sqrt(0.8), 0.9, 0.5, 3/2, 3...
I tried the maximum to be:
3, 3/2, 2, 1, 9/8, sqrt(3)/2...

None are correct. This is a picture of the graph.

Any help would be much appreciated.

Answer 3

Hitaro9, I asked the same exact question. I'm assuming "x+2y" is the overall equation.

Answer 4

Hmm. Is this being given from some website or was it a problem your professor wrote out, because it might be that they just made an error?

Answer 5

This is from "WebAssign."

Unfortunately, this material is from months back, and he gave us 90 more tries on any problems we couldn't answer before final grades were posted. And this was one of the few I couldn't answer, haha.

Answer 6

Yeah. I'm not entirely sure. I'm sorry.

Answer 7

No problem, thank you for your help!

Answer 8

Ask in the chat, there are people way smarter than I am there

Answer 9

And if that doesn't work, try yahoo answers, they've got some smart people too.

Answer 10

Haha so be it!