I understand how to find the largest and smallest values for all my other problems except for this one. The less than or equal to sign is where i'm really getting stuck. Can someone please explain to me how i would treat this equation.
The furthest I've gotten is (x-1/2)^2+y^2=-K+1/4. I don't understand how to treat the x^2+y^2<=1
Question : Find the largest and smallest value of f(x,y) = x - x^2 -y^2 if x^2 +y^2 <= 1
The furthest I've gotten is (x-1/2)^2+y^2=-K+1/4. I don't understand how to treat the x^2+y^2<=1
Question : Find the largest and smallest value of f(x,y) = x - x^2 -y^2 if x^2 +y^2 <= 1
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Look at this geometrically.
x^2 y^2 <= 1 means (x,y) is on or inside the unit circle centered at the origin. In particular, note that |x| and |y| are both <=1
f(x,y) = x - (x^2 y^2) is x minus the distance^2 from (x,y) to the origin (0,0). The minimum is easy. Minimize x (-1) and maxmize the distance from the origin (1), and you get -2, at point (-1,0)
The maximum is trickier. I'm not too good at multivariate calc, not having studied that in over 40 years, but if you take the partial derivatives, and find where they are both 0, that may work. If there's a point where they are both 0, that may give you an extremum.
Find where df/dx =0 and df/dy=0
df/dx = 1 - 2x
df/dy = -2y
The maximum should be where x = 1/2 and y = 0
f(1/2, 0) = 1/2 - 1/4 - 0 = 0.25
Let's look at points close to that
f(1/2 h, 0 k) = 1/2 h - (1/2 h)^2 - k^2
= 1/2 h - 1/4 - h - h^2 - k^2 = 1/4 - h^2 - k^2
Since we are looking for the max, this shows we've found it.
h^2 and k^2 are both >=0, so 1/4 - h^2 - k^2 will be smaller if h or k are non-zero.
I used Solver in Excel to look for the minimum, and it gave me the same answer, (x, y) = (1/2, 0)
x^2 y^2 <= 1 means (x,y) is on or inside the unit circle centered at the origin. In particular, note that |x| and |y| are both <=1
f(x,y) = x - (x^2 y^2) is x minus the distance^2 from (x,y) to the origin (0,0). The minimum is easy. Minimize x (-1) and maxmize the distance from the origin (1), and you get -2, at point (-1,0)
The maximum is trickier. I'm not too good at multivariate calc, not having studied that in over 40 years, but if you take the partial derivatives, and find where they are both 0, that may work. If there's a point where they are both 0, that may give you an extremum.
Find where df/dx =0 and df/dy=0
df/dx = 1 - 2x
df/dy = -2y
The maximum should be where x = 1/2 and y = 0
f(1/2, 0) = 1/2 - 1/4 - 0 = 0.25
Let's look at points close to that
f(1/2 h, 0 k) = 1/2 h - (1/2 h)^2 - k^2
= 1/2 h - 1/4 - h - h^2 - k^2 = 1/4 - h^2 - k^2
Since we are looking for the max, this shows we've found it.
h^2 and k^2 are both >=0, so 1/4 - h^2 - k^2 will be smaller if h or k are non-zero.
I used Solver in Excel to look for the minimum, and it gave me the same answer, (x, y) = (1/2, 0)