minimum using Lagrange multipliers I need to find the minimum x,y, and z coordin

Question

minimum using Lagrange multipliers I need to find the minimum x,y, and z coordinates of the paraboloid x^2+y^2=z with the constraint 6x-6y+108=36z. Please show me all the steps

Explanation / Answer

We want to optimize f(x,y,z)= x^4 + y^4 + z^4 subject to the constraint g(x,y,z) = x^2 + y^2 + z^2 = 1. By Lagrange Multipliers, ?f = ??g ==> = ?. Equating like entries: 4x^3 = 2?x ==> x(2x^2 - ?) = 0 4y^3 = 2?y ==> y(2y^2 - ?) = 0 4y^3 = 2?z ==> z(2z^2 - ?) = 0. (i) Note that x = y = z = 0 can't occur, as it violates the constraint. (ii) Exactly two of the variables equal 0. Say x = y = 0; then the constraint yields z = ± 1. (Similarly for the other two possibilities...) ==> (x, y, z) = (0, 0, ± 1). (0, ± 1, 0), (± 1, 0, 0). Note that f = 1 at these points. ------------- (iii) Exactly one of the variables equals 0. Suppose z = 0 (the other case are treated similarly). We want to optimize f(x,y,0)= x^4 + y^4 subject to g(x,y,0) = x^2 + y^2 = 1. By Lagrange Multipliers, ?f = ??g ==> = ?. Equating like entries (since x, y are nonzero): 4x^3 = 2?x ==> 2x^2 - ? = 0 4y^3 = 2?y ==> 2y^2 - ? = 0 So, ?/2 = x^2 = y^2 Plugging this into g yields ?/2 + ?/2 = 1 ==> ? = 1. So, (x, y, z) = (±1/v2, ±1/v2, 0) [eight sign possibilities]. (similarly for other cases.) At these points, f = 1/2. ----------- (iv) None of x, y, z = 0. So, ?/2 = x^2 = y^2 = z^2. Plugging this into g yields ?/2 + ?/2 + ?/2 = 1 ==> ? = 2/3. So, (x, y, z) = (±1/v3, ±1/v3, ±1/v3) [eight sign possibilities]. At these points, f = 1/3 ------------------- In summary, the maximum value is 1, and the minimum value is 1/3. (Are you sure about the maximum value being v3 ?)

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