F(x,y) = 40x - x^{2} + 15y - 4y^{2 }How do I know what this function is supposed to look like without drawing it out on paper?
Any help would be greatly appreciated
Hey mathbeginner97.
Usually what happens is that you calculate the when the first derivatives are zero and when the function is also zero. In other words find when f(x) = 0 and when f'(x) = 0.
Once you do that you draw a smoioth line that goes through these points to get an idea of what the curve looks like.
Have you done calculus yet (i.e. differentiation)?
For this particular function, $\displaystyle F(x,y) = 40x - x^2 + 15y - 4y^2$, complete the square:
$\displaystyle F(x, y)= -(x^2- 40x)- 4(y^2- (15/4)y)= -(x^2- 40x+ 400- 400)- 4(y^2- (15/4)y+ (225/64)- (225/64))$
$\displaystyle = -(x- 20)^2+ 400- 4(y- 15/8)^2+ 225/8= 3425/8 -(x- 20)^2- 4(y- 15/8)^2$
That's a "paraboloid" with vertex at (20, 15/8, 3425/8) and axis the line x= 20, y= 15/8, z= t.
Answering your question more generally is exactly what "multi-variable calculus" is for.