MHB JOHN's question at Yahoo Answers involving Lagrange multipliers

[MarkFL]

Here is the question:

Calc 3 Lagrange multiplier question?

What is the maximal cross-sectional area of a rectangular beam cut from an elliptical log with semiaxis of length 2 feet and 1 foot?

I need help on applying the method of lagrange to this scenario. Thank you

Here is a link to the question:

Calc 3 Lagrange multiplier question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello JOHN,

I would choose to orient my coordinate axes such that the origin coincides with the center of the elliptical cross-section of the log, where all linear measures are in ft. Now looking only at the first quadrant, let's let the upper right vertex of the cross section of the rectangular beam be at $(x,y)$. Hence, adding the other 3 quadrants, we find the area of this rectangular cross-section, i.e., our objective function is:

$$A(x,y)=4xy$$

subject to the constraint:

$$g(x,y)=x^2+4y^2-4=0$$

Using Lagrange multipliers, we obtain the system:

$$4y=\lambda(2x)$$

$$4x=\lambda(8y)$$

Solving both for $\lambda$, and equating, we find:

$$\lambda=\frac{4y}{2x}=\frac{4x}{8y}$$

Simplifying, we find:

$$x^2=4y^2$$

Substituting this into the constraint, we find:

$$4y^2+4y^2=4$$

Solving for $y^2$, we find:

$$y^2=\frac{1}{2}$$

and since:

$$x=2y$$, as we have taken the positive root given the two variables are in the first quadrant, we find:

$$A_{\text{max}}=A(2y,y)=8y^2=8\cdot\frac{1}{2}=4$$

Hence, the maximal area of the rectangular cross-section of the beam is $4\text{ ft}^2$.

To JOHN and any other guest viewing this topic, I invite and encourage you to post other optimization with constraint problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.