Question

a wire 34 cm long is cut into 2 pieces. One piece is bent to form a square and the other is bent to form a rectangle that is twice as wide as it is long. How should the wire be cut in order to minimize the total area of the square and the rectangle?

Answer 1

start with an equation of lengths, the rectangle will be of length A, the Square length B: Therefore A+B=34. Then make an equation for the area of the square. Each side is length (B/4), the area= (length)x(Width) therefore the area is (B/4)^2 aka (1/16)B^2. now you need an equation for the area of the rectangle. So divide that length of A into 6 segments, 2 segments will make each long side of the and 1 for each end. The area of that rectangle is ((2 segments)(A/6))*(1 segment)(A/6) simplify to (A/3)(A/6) simplify once more to (1/18)A^2. Now you have a system of equations, but need to relate them all together. Looking back to the question you need to maximize the area of each, so you could say that you need to maximize the area of the sum. So taking Total area= (1/16)B^2 + (1/18)A^2 you need to find the maximum of that equation. Plug in the relationship of A to B, aka A= 34-B. now you have (1/16)B^2 +(1/18)(34-B)^2= total area. Now find when the derivative is zero. d(total area)/d(length B) = (1/8)B + ((-34/9)+(1/9)B) find the zero of that. B=16 should be the zero. finally plug back in to our first equation and you can see that B=16 and A should be 18. Hopefully that helps, here's just a bit more information, the area of the square is 16, the area of the rectangle should be 18.

Answer 2

I should have also stated, that you need to find the zero when the rate of change goes from negative to positive, because it would represent a decreasing area, then a zero or minimum area, then an increasing area. As a check you could just guess the length of the square to be B=10, then the area of the square should be 6.25, the length of A would then be 24, which gives a area of the rectangle of 32. Add them together for the total, you see that you would have 38.25 square cm. This is higher than the minimal 34 square cm that we calculated above. therefore we are right and can prove that our work makes sense.

Answer 3

thank you so much!