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Find max value on the border of f[x,y]

Posted 4 years ago

We are instructed to make a parametric expression of the circular border of an f[x,y] function:

g[t_]:=f[R*cos[t],R*Sin[t]]

I don't understand the connection between the function and the parametric expression. The function is defined as

f[x_, y_] := 10 Log[x^2 + 2 y^2 + 1] - x^2 - y^2;

how does Mathematica "understand" that I am referring to the border of this particular function when I write the parametric expression of the border?

The g[t_] has got neither "x" nor "y" in it so how does Mathematica understand that I am making an expression on the values of these specific variables at the circular border?

POSTED BY: Peter Lahti
3 Replies

I interpret your question as find the extrema of the function f[x,y] where x and y lie on some curve. In that case:

    f[x_, y_] := 10 Log[x^2 + 2 y^2 + 1] - x^2 - y^2
    Plot[f[Cos[t], Sin[t]], {t, 0, 2 Pi}]
    Plot3D[f[x, y], {x, -2, 2}, {y, -2, 2},   RegionFunction -> ((#1^2 + #2^2 < 1) &)]
    Solve[D[f[Cos[t], Sin[t]], t] == 0, t]

or in the g[t] form

g[t_] := f[{x[t], y[t]}

where {x[t],y[t]} is your curve.

POSTED BY: W. Craig Carter
Posted 4 years ago

The domain of the values that the variables of the function "is allowed" to have is a radius of 4. x can be -4 to 4, y -4 to 4. So the area is a circle. I am supposed to find the maximum value of the function at the circular border of the function (so this value is neither a stationary point nor a max or min value for the whole function, it is just the max of the border of the domain). I understand that one step towards achieving this, is to make a parametric expression of the border: g[t_]=RCos[t]+RSin[t], where R=4. (must be, right?) But I do not understand the connection between f[x,y] and g[t]. How do I get Mathematica to "understand" that I am writing a parametric expression for the border line of the function?

POSTED BY: Peter Lahti
Posted 4 years ago

Hi Peter,

What do you mean by "circular border of an f[x,y] function"?

POSTED BY: Rohit Namjoshi
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