Optimizations Technique

[371204561]

For a rectangle whose perimeter is 20 m, use the Lagrange multiplier method to find the
dimensions that will maximize the area.

[Peter Pedant]

you really must do your own homework
it is set on what you have done during the week

anyway
https://en.wikipedia.org/wiki/Lagrange_multiplier#Examples

look at ex 1 - instead of a circle - you change it to a rectangle ....

[jim360]

To start you off, how do we work out the area A and perimeter P of a rectangle?

The Lagrange multiplier technique spits these into a function A - l (P-20), where l is the multiplier, and asks us to differentiate this with respect to x,y and also l.

It should look like:

A = xy
P = 2(x+y) = 20

A- l (P-20) = xy - 2l (x+y-10)

Hopefully you know how to differentiate this with respect to x and y etc.

[fiction-factory]

Hi. Were our answers helpful on your previous question last week as you never responded then. Yse Jim's method here and this is straightforward. you don't need to know what a lagrange multiplier is _ I have forgotten but can still solve this.

[fiction-factory]

Anyway, without even writing anything down I'd say intuitively the maximum area of a rectangle of given perimeter has to be when its a square

[fiction-factory]

9 x1 =9
7 ×3 =21
6 ×4 =24
5×5 =25

[fiction-factory]

I assumed above that the answer has to be a rectangle (which can include a square). If any shape is allowed then the answer is a circle so your answer will have reciprocal of pi in it

[fiction-factory]

Without using Lagrange multiplier method I'd have simply done it this way:

Let width=x and length =y
Area (A) = xy
Perimeter = 2y +2x= 20, so y+x =10 so y= 10-x
So A = x(10-x)= 10x-x²
To find the min/max: dA/dx= 10-2x which is 0 at the max/min , i.e. when x=5
Second derivative is negative (-2) so it's a maximum

When x=5, y=10-5=5
So answer is 5m x 5m