Chapter 14.5, Problem 11E

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Chapter
Section

### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Find the minimum value of w   =   x 2 +   y 2 +   z 2 subject to the constraint x   +   y   +   z =   3 .

To determine

To calculate: The minimum value of w=x2+y2+z2, which is subjected to the constraint x+y+z=3.

Explanation

Given Information:

The provided function is w=x2+y2+z2 and it is subjected to the constraint x+y+z=3.

Formula used:

Lagrange Multipliers Method:

According to the Lagrange multipliers method to obtain maxima or minima for a function w=f(x,y,z) subject to the constraint g(x,y,z)=0,

Step 1: Find the critical values of f(x,y,z) using the new variable λ to form the objective function F(x,y,z,λ)=f(x,y,z)+λg(x,y,z).

Step 2: The critical points of f(x,y,z) are the critical values of F(x,y,z,λ) which satisfies g(x,y,z)=0.

Step 3: The critical points of F(x,y,z,λ) are the points that satisfy:

Fx=0, Fy=0, Fz=0 and Fλ=0, that is, the points which make all the partial derivatives equal to zero.

For any function f(x,y,z), the partial derivative of f(x,y,z) with respect to y is calculated by taking the derivative of f(x,y,z) with respect to y and keeping the other variables x and z constant.

The partial derivative of f(x,y,z) with respect to y is denoted by fy.

And, similarly the partial derivatives of function f(x,y,z) with respect to x and z can be calculated as above.

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Constant function rule for a constant c is such that, if f(x)=c then f(x)=0.

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the function, w=x2+y2+z2.

The provided constraint is x+y+z=3.

According to the Lagrange multipliers method,

The objective function is F(x,y,z,λ)=f(x,y,z)+λg(x,y,z).

Here, f(x,y,z)=x2+y2+z2 and g(x,y,z)=x+y+z3.

Put the values of f(x,y,z)=x2+y2+z2 and g(x,y,z)=x+y+z3 in the objective function, F(x,y,z,λ)=f(x,y,z)+λg(x,y,z)

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