MHB Holly's questions at Yahoo Answers regarding Lagrange multipliers

[MarkFL]

Here are the questions:

Calculus 3 Lagrange multipliers help?


Okay I CANNOT figure out lagrange multipliers. Can anyone help me with my homework problems? Thanks!

1.

Use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive.
minimize f(x,y)= x^2 + y ^2
constraint: x+ 2y - 20 = 0

2.
Use Lagrange multipliers to find the indicated extrema, assuming that x and y are positive.

Maximize f(x,y) = (99 - x^2 - y^2)^(1/2)
constraint: x + y - 10 = 0

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Holly,

1.) We are given the objective function:

$$f(x,y)=x^2+y^2$$

subject to the constraint:

$$g(x,y)=x+2y-20=0$$

Lagrange multipliers gives rise then to the following system:

$$2x=\lambda(1)$$

$$2y=\lambda(2)$$

This implies:

$$\lambda=2x=y$$

Substituting for $y$ into the constraint, we find:

$$x+2(2x)-20=0\implies x=4\implies y=8$$

Thus, we obtain the critical point:

$$(x,y)=(4,8)$$

The objective function's value at this point is:

$$f(4,8)=4^2+8^2=80$$

To ensure this is a minimum, let's evaluate the objective function at another point on the constraint, such as $$(x,y)=(2,9)$$:

$$f(2,9)=2^2+9^2=85$$

And so we may now conclude:

$$f_{\min}=f(4,8)=80$$

2.) We are given the objective function:

$$f(x,y)=\sqrt{99-x^2-y^2}$$

subject to the constraint:

$$g(x,y)=x+y-10=0$$

Now here we see that $x$ and $y$ have cyclic symmetry, that is, we may switch the two variables and still have the same objective function and constraint. Thus we know the critical value may be obtained when $x=y$. The constraint then gives us:

$$x=y=5$$

And so our critical point is:

$$(x,y)=(5,5)$$

The objective function's value at this point is:

$$f(5,5)=\sqrt{99-5^2-5^2}=7$$

To ensure this is a maximum, let's evaluate the objective function at another point on the constraint, such as $$(x,y)=(4,6)$$:

$$f(5,5)=\sqrt{99-4^2-6^2}=\sqrt{47}<7$$

And so we may conclude:

$$f_{\max}=f(5,5)=7$$