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An iron wire $3$ meters long is cut in two. We form a square with the first piece and an equilateral triangle with the second. How must it be cut for the total area of these two figures to be maximized? Round the answer to two decimal places.

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1 Answer 1

Let the two pieces be of length $x$ meters and $3-x$ meters.

The perimeter of the square is $x$ meters and the perimeter of the equilateral triangle is $3-x$ meters.

Hence, the side of the square is $x/4$ meters and the side of the equilateral triangle is $\dfrac{3-x}3 = 1 - \dfrac{x}3$ meters.

The area of the square is $(x/4)^2$ and the area of the equilateral triangle is $\dfrac{\sqrt{3}}4\left( 1 - \dfrac{x}3\right)^2$.

Hence, the total area is $$A(x) = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}4\left( 1 - \dfrac{x}3\right)^2$$ Now use calculus to maximize this function given that $x \in [0,3]$.

Move your mouse over the gray area for the final answer.

$$\dfrac{dA}{dx} = \dfrac{x}8 + \dfrac{\sqrt{3}}2 \left(\dfrac{x}3-1 \right) = 0 $$ This gives us $$x = \dfrac{12}{13} (4-\sqrt{3}) \approx 2.093491562244113267513126 \approx 2.09 m$$ This gives us the minimum since the second derivative is $>0$. The maximum hence occurs on the boundary i.e. either at $x=0$ or $x=3$. $$A(0) = \dfrac{\sqrt{3}}4 \approx 0.4330127018922193233818615853764680917357013134525951$$ $$A(3) = 0.5625$$ Hence, to maximize the area use the entire wire to make a square.

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That was not very pedagogical. –  tst Nov 24 '12 at 3:52
    
Answer is not correct –  Tobi Nov 24 '12 at 3:58

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