Find the absolute maximum and the absolute minimum of the function f (x, y) = x^

Question

Find the absolute maximum and the absolute minimum of the function
f (x, y) = x^ 2 + xy - y
on the closed triangle with vertices at the points (0, 0), (0,

Explanation / Answer

Our region is bounded by x = 0, y = -3, x = -y dz/dy = x - 1 = 0 => x = 1.....Put this in dz/dx dz/dx = 2x + y =0 => y = -2 So at (1,-2), z=1 Now we need to check at the boundaries. f = z(0, y) = -y => df/dy = -1 This means that the function is decreasing. So the critical points on this boundary are (0,0) and (0,-3) z (0,0) = 0 and z(0,-3) = 3 Now on the boundary y =-3, f = z(x,-3) = x^2 -3x +3 => df/dx = 2x-3 => x = 1.5 So (1.5,-3) is a critical point along with the corners i.e. (0,-3) and (3,-3) z(1.5,-3) = 0.75 z(0,-3) = 3 z(3,-3) = 3 Now checking at x = -y, f = z(-y, y) = -y =>df/dy = -1 Again a decreasing function so the critical points are again the corners (0,0) and (3,-3). We have already checked these points. So finally, there are two absolute maxima, (0,-3) and (3,-3) where z = 3 and one absolute minimum, (0,0) where z = 0

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.