Verified answer

f = x² - y²

g = x - 6y + 105 = 0

∇ f = λ ∇ g

2x = λ

2y = 6 λ

⇒ y = 6x

x = 3 and y = 18 ... plug y = 6x into g

(3,18) is a minimum, there is no global maximum

Answer: (3,18) is a min, no global max

i will provide it a attempt, yet i do no longer comprehend if my answer would be appropriate. 2xy^2z^2 = 2x? 2yx^2z^2 = 2y? 2zx^2y^2 = 2z? x^2+y^2+z^2 = a million xy^2z^2 = x? yx^2z^2 = y? zx^2y^2 = z? x^2+y^2+z^2 = a million Multiply equation a million by utilizing x, and equation 2 by utilizing y xy^2z^2 = x? x^2y^2z^2 = x^2? yx^2z^2 = y? x^2y^2z^2 = y^2? Subtract the recent equations 0 = x^2? - y^2? element out ? 0 = ?(x^2 - y^2) it relatively is not any longer correct what ? is, considering that we are able to divide that out. If ? = 0, then x,y,z = 0, which does no longer fulfill the regulations. So 0 = ?(x^2 - y^2) = (x+y)(x-y) Giving x = -y, and x = y Multiply equation 2 by utilizing y, and equation 3 by utilizing z and subtract, giving y^2? - z^2? = 0 ?(y^2 - z^2) = 0 ?(y-z)(y+z) = 0 So y = z, and y = -z ---> y = z, -y = z And x = -y = z, and x = -y = z. Plugging in... x^2 + y^2 + z^2 = a million x^2 + (-x)^2 + x^2 = a million 3x^2 = a million x^2 = a million/3 x = (a million/3)^(a million/2), and -(a million/3)^(a million/2) and so on. Edit: pass with kb's answer, he's helped me plenty with my own calc questions!