A piece of wire $26 m$ $26m$ 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 minimize the total area?

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A piece of wire $26 m$ $26m$ 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 minimize the total area?

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Answer 1 · 2018-02-22T23:36:10

$11.309 c m$ $11.309 cm$

Explanation:

Let $y$ $y$ equal the length of side of triangle and $x$ $x$ equal length of side of square.

So the perimeter of both the triangle and square will equal $3 y + 4 x = 26$ $3y+4x=26$ ,............. $[1]$ $[1]$

The area of the triangle and square combined will be

$\frac{\sqrt{3}}{4} y^{2} + x^{2}$ $sqrt3/4y^2+x^2$ ,............ $[2]$ $[2]$ ,the first term is the area of an equilateral triangle of side length $y$ $y$ [from Pythagoras]
From $[1], x = \frac{26 - 3 y}{4}$ $[1] , x=[26-3y]/4$ and substituting this value for $x$ $x$ into.. $[2]$ $[2]$ we obtain,

Total area $= sqrt3/4y^2+[[26-3y]/[4]]^2...............[3]$

Differentiating $[3]$ using the chain rule, $2sqrt3/4y + 1/16d/dy[26-3y]^2$ . $d/dy [26-3y]^2=2[26-3y][-3]= -6[26-3y]$ and so we have ,

$Da/[dy]=2sqrt3/4y-6/16[26-3y]=0$ ,for max or min .

Tidying up , $4sqrt3y-78+9y =0,............$ 4], and factoring out #y#,

and so $y$ = $78/[4sqrt3+9$ $= 4.8969$ , so perimeter of triangle = $3[ 4.8969]$
= $14.6907 cm$ and so the perimeter

the square is $26-14.6907=11.309 cm$ .

From............. $[4],$ $d^2A/dy^2= 4sqrt3+ 9 -78$ which is negative and so this shows that the value of $y$ found will maximise the area.

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A piece of wire $26 m$ $26m$ 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 minimize the total area?

1 Answer

Explanation:

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A piece of wire 26m26m26m 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 minimize the total area?

1 Answer

Explanation:

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A piece of wire $26 m$ $26m$ 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 minimize the total area?