Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, please don't yell at us. It's rude. – Cameron Buie Nov 24 '12 at 3:53

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.

share|cite|improve this answer
That was not very pedagogical. – tst Nov 24 '12 at 3:52
Answer is not correct – Tobi Nov 24 '12 at 3:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.