simple question for the maths retard like me. (1 Viewer)

[kuromusha]

I have a simple Cobb Douglas function. 5 = xy

which is y = 5/x

On graph it is concave from the origin in the first quadrant, how do I flip it so that it is convex from the origin?

ie how do I turn do this

Is that even possible or am I making a fool out myself?

reason, trying to study Edgeworth box without flipping my screen around.

 

[si2136]

For your second graph, does it approach +infinity on both axis?

 

[InteGrand]

I have a simple Cobb Douglas function. 5 = xy

which is y = 5/x

On graph it is concave from the origin in the first quadrant, how do I flip it so that it is convex from the origin?

ie how do I turn do this

Is that even possible or am I making a fool out myself?

reason, trying to study Edgeworth box without flipping my screen around.

Do you actually need to flip that curve? Or would any concave function suffice? Or do you need certain properties?

Even if we wanted to flip that curve, there's more than one way to do it. To flip it how you want, you'd want to reflect it about a line of the form y = -x + b for some b > 0. You'd need to specify b to find exactly what flip it'll be (or tell us any conditions needed for the curve, which may help specify the b).

One simple option if you don't specifically need to flip that branch of the hyperbola is to translate the other branch of the hyperbola along the diagonal line y = x. This second hyperbola-branch has equation y = 5/x, x, y < 0. To translate this by a units up and right, the new equation is just:

(x – a)(y – a) = 5, x, y < a (a > 0 is the amount of vertical and horizontal shift).

If you only want to consider the part of this curve in the first quadrant, you'd include the constraints x, y > 0.