# What is the maximum value of the function f(x,y,z)=x+2y+3z,on the curve on intersection on the plane x-y+z=1 and cylinder x^2+y^2-1=0?

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### 1 Answer

The function is subject to 2 constraints: g(x,y,z) = x-y+z=1 and h(x,y,z) = x^2+y^2-1=0.

We'll use the Lagrange's five equations to determine the maximum value of the function.

df/dx = s*(dg/dx) + t*(dh/dx)

1= s + 2x*t (1)

df/dy = s*(dg/dy) + t*(dh/dy)

2 = -s + 2y*t (2)

df/dz = s*(dg/dz) + t*(dh/dz)

3 = s (3)

x - y + z = 1 (4)

x^2+y^2 = 1 (5)

We'll substitute s = 3 in (1):

s + 2x*t = 1

3 + 2x*t = 1

2x*t = -2

x*t = -1

x = -1/t (6)

We'll substitute s = 3 in (2):

2 = -3 + 2y*t

2yt = 5

y = 5/2t (7)

We'll substitute (6) and (7) in (5):

(1/t^2) + (25/4t^2) = 1

29/4t^2 = 1

t^2 = 29/4

t1 = -sqrt29/2 ; t2 = sqrt29/2

x1 = 2/sqrt29 ; x2 = -2/sqrt29

y1 = -5/sqrt29 ; y2 = 5/sqrt29

z1 = 1 - 7/sqrt29 ; z2 = 1 + 7/sqrt29

The maximum value of f is:

f max = -2/sqrt29 + 2*5/sqrt29 + 3(1+7/sqrt29)

fmax = -2/sqrt29 + 10/sqrt29 + 3 + 21/sqrt29

fmax = 29/sqrt29 + 3

fmax = 29*sqrt29/29 + 3

fmax = sqrt29 + 3

**The maximum value of f is: fmax = sqrt29 + 3.**