The sum of the perimeters of an equilateral triangle and a square is 10. Find the dimensions of the triangle and the square that produce a minimum total area.

x - side of an equilateral triangle, y - side of a square;
3 x + 4 y = 10
4 y = 10 - 3 x
b = ( 10 - 3 x ) /4
Total area:
A = x ²√3/4 +y² = x²√3/4 + ( 10- 3 x )²/16 = x²√3/4 + 9 x²/16 - 15 x/4 + 25/4
A= x√3/2 + 9 x/8 - 15/4 A = 0
4 x√3 + 9 x = 30
15.92 x = 30
x = 30 : 14.92 ≈ 1.88
y = (10 - 1.88 ) : 4 ≈ 2.03
Answer: dimensions are x = 1.88 ( triangle ) and y = 2.03 ( square )

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calculista calculista · Answer 2 · 2018-08-02T14:24:56-04:00

Let

x-------> the length side of the equilateral triangle

y-------> the length side of the square

we know that

The sum of the perimeters of an equilateral triangle and a square is \(10\)

Perimeter of triangle is equal to \(3x\)

Perimeter of the square is equal to \(4y\)

so

\(3x+4y=10\\ 4y=10-3x\)

\(y=2.5-0.75x\) ------> equation \(1\)

Find the area of equilateral triangle

Applying the law of sines

\(A1=\frac{1}{2} *x^{2} *sin 60\\ A1=\frac{\sqrt{3}}{4} *x^{2}\)

Find the area of the square

\(A2=y^{2}\)

Fin the total area

\(At=A1+A2\)

\(At=\frac{\sqrt{3}}{4} x^{2} +y^{2}\) ----> equation \(2\)

Substitute equation \(1\) in equation \(2\)

\(At=\frac{\sqrt{3}}{4} x^{2} +(2.5-0.75x)^{2}\)

Using a graph tool

see the attached figure

we know that

the vertex of the graph is the point with the minimum total area

the vertex of the graph is the point \((1.88,2.72)\)

that means that

for \(x=1.88 units\) the total area is equal to\(2.72 units^{2}\) (is the minimum total area)

find the value of y

\(y=2.5-0.75*1.88\)

\(y=1.09 units\)

therefore

the answer is

the length side of the equilateral triangle is equal to \(1.88 units\)

the length side of the square is equal to \(1.09 units\)