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| Math Central | Quandaries & Queries |
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Question from Raahim: A 2 meter piece of wire is cut into two pieces and one piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire cut so that the total area enclosed by both is minimum and maximum? |
Hi,
Suppose that when you cut the wire one of the pieces is $x$ meters lang then the other is $2 - x$ meters long. You know that $0 \leq x \leq 2.$
Use the piece of length $2 - x$ meters to form the square. Each side of the square is then $\large \frac{2-x}{4}$ meters long. What is the area of the square?
Use the piece of length $x$ meters to for the equilateral triangle and hence each side of the triangle will be $\large \frac{x}{3}$ meters long. Use the technique described in my response to Ginny to calculate the area of the triangle.
Add the area of the square and the area of the triangle and use your knowledge of calculus to complete the problem.
Penny
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