Calculate absolute maxima and minima of a two variable function

I have a function

% f(x,y) = x^4 + y^4 - x^2 - y^2 + 1

but i am not able to understand how to start solving this. Can someone help me with the code?

Answer 1

[Image Analyst]

OK, here is the function, but exactly what does "solve" mean to you???

xMin = -2;
xMax = 2;
yMin = -2;
yMax = 2;
numPoints = 200;
xv = linspace(xMin, xMax, numPoints);
yv = linspace(yMin, yMax, numPoints);
[x, y] = meshgrid(xv, yv);
% f(x,y) = x^4 + y^4 - x^2 - y^2 + 1
fprintf('Creating function.\n');
f = x.^4 + y.^4 - x.^2 - y.^2 + 1;
fprintf('Creating surface plot.\n');
surf(x, y, f, 'LineStyle', 'none');
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);
zlabel('f', 'FontSize', 20);
title('f(x,y) = x^4 + y^4 - x^2 - y^2 + 1', 'FontSize', 20);
colorbar;

Do you want to use contour() or contour3() to find out where it equals some value?

[Rishabh Agrawal]

For me, solve means to display the values of absolute maxima and minima, but i am understanding your solution a little bit, thanks for your help

[Image Analyst]

maxValue = max(abs(f(:)))
minValue = min(abs(f(:)))
fprintf('The max of f = %f.\nThe min of f = %f.\n', maxValue, minValue);

If this does what you wanted, then please "Accept this answer".

[Image Analyst]

Note: that max is for the plotted region. If you plotted more, the max would be higher. For x and y of infinity, the max is infinity.

The min though is always at (x,y) = (0,0) and is 1.

Answer 2

[Billuri]

f1 = x^2 + y^2

[Image Analyst]

Can you please elaborate on how this solves his question on the 4th order polynomial? He says "solve means to display the values of absolute maxima and minima".