Question

Vaidehi wants to cut a 30-meter piece of iron into two pieces. One of the pieces will be used to build an equilateral triangle, and the other to build a rectangle whose length is three times its width. Where should Vaidehi cut the iron bar if the combined area of the triangle and the rectangle is to be a minimum? How could the combined area of these two figures be a maximum? Justify your answers.

How does one create the equation for something like this?

Answer 1

I got you, just a sec

Answer 2

Another optimization problem. It took a little time for me to think it through tho damn.

1. First collect the formulas and equations you can use:
\( A_{equilateral~triangle} = \frac{√3}{4}s^2\)
\( A_{rectangle} =lw = 3w(w) = 3w^2 \)
\( net~Area = A_{rectangle} + A_{equilateral~triangle} = 3w^2 + \frac{√3}{4}s^2 \)

Now common sense would be that whatever length of the piece used for the Area of the rectangle would be 30 minus the area of the triangle. Mathematically, we could say it like this:

\(s = 30 - w\)

or to make things easier in the future calculations..

\(w = 30 - s\)

Now substitute

\( net~Area = A(s) = 3(30-s)^2 + \frac{√3}{4}s^2 \)

Now you derivate this thing and I'm just gonna do that on the calculator because otherwise it would take toooo longg

\( \frac{d}{ds} A(s) = 6s + \frac{√3}{2}s - 180 \)
Find your relative extrema by setting to 0
s = 26.216

We get our extremum as 26.216 m.

There's no need to plug it into the second derivative to see if it's a minimum or maximum because that's the only extremum and knowing that the area function is a lower bound quadratic, we can say it is a minimum. So your answer would be 26.216 m to minimize the net area.

Answer 3

I think the only way to find the maximum area is to put a bound on the function by restricting it because the max amount of \( s \) or \( w \) is 30.

Answer 4

Since our function A(s) is with respect to the side of the equilateral triangle, we can restrict it by a domain of 0≤s30 and then find the critical points.

Answer 5

The maximum Area would be when \( s = 0 \) and \( w = 30 \) based on the net area function \( net~Area = A(s) = 3(30-s)^2 + \frac{√3}{4}s^2 \) including the domain restriction of \( 0 ≤ s ≤ 30 \)

I graphed it and then you'll see why
https://www.desmos.com/calculator/lfflsfdx0d

Answer 6

you need to refine this!

3s + 8w = 30

not s + w = 30

yep?

And then another step for the actual cut.

Answer 7

I didn't think of that @SemiDefinite thanks for informing. I am going through this concept myself so I do expect a few discrepancies here and there, but your helpful advice will come handy

So like Semi said, you'll probably need to solve for w from 3s + 8w = 30 and substitute that into the net area function and differentiate it all over again.