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# A piece of wire 5 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

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A piece of wire 5 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

How much wire should be used for the square in order to maximize the total area?

How much wire should be used for the square in order to minimize the total area?

Nov 20, 2018

#1
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This is not exactly worded correctly. Technically, the square area has a maximum area of $$1.5625m^2$$ as  $$(5/4)^2=1.5625$$ .

The minimum area is $$0m^2$$ because $$(0.0...1)^2=0$$.

You are very welcome!

:P

Nov 21, 2018
#3
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Actually, since the wire is cut into two pieces (that are equal, I assume), the perimeter of the square is 5/2 meters or 250 cm. Each side of the square would then be 250/4 cm, which equals 62.5 cm. $$62.5^2$$  is 3906. The total area of the square would then be 3906.25 $$cm^2$$ or 0.390625 $$m^2$$

The minimum area would be 0...just don't place the wire.

I think you meant to state the question differently...

Hope this sort of helps,

- PartialMathematician

PartialMathematician  Nov 21, 2018
#5
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Ok then... :P

PartialMathematician  Nov 21, 2018
#2
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I assume you mean the total area of the square + the triangle should be maximized

the square's area = l x l = l^2

the triangle's area = sqrt3/4 * a^2

And    a = (5m - 4l)/3

Have to agree with above....if you want to maximize area, you would not put ANY wire in to a triangle...it would all be in the square......UNLESS you have a size for the triangle...or some sort of ratio of square to triangle....there really is no answer to this question. Nov 21, 2018
edited by ElectricPavlov  Nov 21, 2018
#4
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We also do not know if the 5m wire was cut into 2 equal length wires or 2 random wires.

PartialMathematician  Nov 21, 2018
#7
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To use Alan's expression     mmm

I have edited this because I had forgot the half in the area of a triangle I hope I ahve fixed that error and not created any more mistakes :/

A piece of wire 5 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

How much wire should be used for the square in order to maximize the total area?

How much wire should be used for the square in order to minimize the total area?

Let the perimeter of the equilateral triangle be 6x  then the perimeter of the square will be  5-6x    metres

We know that

$$0<6x<5\\ 0 The side length of the triangle is 2x so the AREA will be \(2x\cdot \sqrt3x=2\sqrt3x^2$$

If all of the wire is used for the triangle then x=5/6 and the area will be  $$\color{red}{0.5*}\color{black}{2\sqrt3\cdot \frac{25}{36}}=\frac{25\sqrt3}{36}\approx 1.2m^2$$

The side length of the square is   $$\frac{5-6x}{4}$$  so its AREA will be   $$\left[\frac{5-6x}{4}\right]^2=\frac{25+36x^2-60x}{16}$$

If all the wire is used for the square then x=0 the area will be  area will be  $$\frac{25}{16}=1.5625$$

So the total Area

$$A=\sqrt3x^2+\frac{25+36x^2-60x}{16}\\ A=\sqrt3x^2+2.25x^2-3.75x+\frac{25}{16}\\ \text{This is a concave up parabola so there will only be a minimum, }\\\text{the max will be at an end of domain point.} \\ \frac{dA}{dx}=2\sqrt3x+4.5x-3.75 \\\frac{dA}{dx}=(2\sqrt3+4.5)x-3.75\\ min\;\; when\;\; \frac{dA}{dx}=0 \\ 0=(2\sqrt3+4.5)x-3.75\\ x=\frac{3.75}{(2\sqrt3+4.5)}\\ x\approx 0.47086\\ \text{which is in the required domain}$$

So   in order to minimize total area the this much of the wire must be used for the square.

$$square\;\; perimeter \approx 5-6*0.47086\\ square\;\; perimeter \approx 2.17 metres$$

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So the minimum total area will be when the square gets approx 2.17m of the total 5m

And the maximum total area will be be gained as the the square perimerter appoaches 5

Nov 22, 2018
edited by Melody  Nov 22, 2018
edited by Melody  Nov 22, 2018
#8
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Here is the graph of  Toatal area in terms of x

Remember :

6x metres is cut off for the TRIANGLE

and

5-6x metres is used for the square

(edited to fit with the corrected answer above) Melody  Nov 22, 2018
edited by Melody  Nov 22, 2018
#9
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Here's my take on this: .

Nov 22, 2018
#10
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Thanks Alan,

I had a super stupid error in mine but I have corrected it and I think our answers are the same now Melody  Nov 22, 2018