Use Lagrange multipliers to find the maximum and minimum values of f(x, y, z) = x − 2y + 5z on the sphere x 2 + y 2 + z 2 = 30.

Answer:Maximum: ((1,-2,5) ; 30)Minimum: ((-1,2,-5) ; -30)Step-by-step explanation:We have the function f(x,y,z) = x - 2y + 5z, with the constraint g(x,y,z) = 30, with g(x,y,z) = x²+y²+z². The Lagrange multipliers Theorem states that, the points (xo,yo,zo) of the sphere where the function takes its extreme values  should satisfy this equation:grad(f) (xo,yo,zo) = λ * grad(g) (xo,yo,zo)for a certain real number λ. The gradient of f evaluated on a point (x,y,z) has in its coordinates the values of the partial derivates of f evaluated on (x,y,z). The partial derivates can be calculated by taking the derivate of the function by the respective variable, treating the other variables as if they were constants.Thus, for example, fx (x,y,z) = d/dx x-2y+5z = 1, because we treat -2y and 5z as constant expressions, and the partial derivate on those terms is therefore 0. We calculate the partial derivates of both f and gfx(x,y,z) = 1fy(x,y,z) = -2fz(x,y,z) = 5gx(x,y,z) = 2x (remember that y² and z² are treated as constants)gy(x,y,z) = 2ygz(x,y,z) = 2zThus, for a critical point (x,y,z) we have this restrictions:1 = λ 2x-2 = λ 2y5 = λ 2zx²+y²+z² = 30The last equation is just the constraint given by g, that (x,y,z) should verify.We can put every variable in function of λ, and we obtain the following equations.x = 1/2λy = -2/2λ = -1/λz = 5/2λNow, we replace those values with the constraint, obtaining(1/2λ)² + (-1/λ)²+(5/2λ)² = 30Developing the squares and taking 1/λ² as common factor, we obtain(1/λ²) * (1/4 + 1 + 25/4) = (1/λ²) * 30/4 = 30 Hence, λ² = 1/4, or, equivalently,[tex]\lambda =^+_- \frac{1}{2} . [/tex]If [tex]\lambda = \frac{1}{2} , [/tex] then 1/λ is 2, and thereforex = 1y = -2z = 5and f(x,y,z) = f(1,-2,5) = 1 -2 * (-2) + 5*5 = 30If [tex]\lambda = - \frac{1}{2} , [/tex] then 1/λ is -2, and we havex = -1y = 2z = -5and f(x,y,z) = f(-1,2,-5) = -1 -2*2 + 5*(-5) = -30.Since the extreme values can be reached only within those two points, we conclude that the maximun value of f in the sphere takes place on ((1,-2,5) ; 30), and the minimun value takes place on ((-1,2,-5) ; -30).