Question

A wire of length L meters is to be divided into two parts; one part will be bent into a square and the other into a circle. How should the wire be divided to make the sum of the areas of the square and circle as large as possible?
Circle ? m of wire
Square? m of wire

How should the wire be divided to make the sum of the areas of the square and circle as small as possible?

Answer 1

You'll want to minimize the number of variables used, or the problem will get unwieldy. So, don't start with something like x + y = L. You'll end up introducing "y" which you don't need. Part of the wire goes to the circle so call that "x". The part of the wire that goes to the square is then "L - x". "L" is fixed, and it will not vary.

Answer 2

Set\[A = A _{s} + A _{c}\] All three are functions of just "x". Formulate the areas of square and circle and add for total area. That is what that equation above represents. Take the first derivative. Then either take the second derivative and check for concavity or just check to see if the first derivative goes from + to - or - to + to check max or min.

Answer 3

Um I'm confused how to set it up especially for the circle since its pir^2

Answer 4

You will have relate "x", for the circle, to circumference and then relate that to a radius and then you can get the area of the circle.

Answer 5

x = (2)(pi)(r), the circumference. r is then x / [(2)(pi)]. Area of circle is (pi)r^2. So in this last formula, you substitute x / [(2)(pi)] for radius and now you have an equation for the area of the circle strictly in one independent variable, "x". The square is easier and that is also only in "x" because "L" is fixed, not variable. Total equation for combined area is in one variable. Easy from that point onward.

Answer 6

For square it is (L-x)^2 now u have an L in there, what do u do

Answer 7

3 things.
First, the area for the square will NOT be (L - x)^2 because L - x is the perimeter of the square, not the length of a side. Given that L - x is the perimeter, you should be able to find the length of any side of the square and then find the correct formula for the area of the square.

Second, as I mentioned already, "L" is a constant, not a variable, so it will present no problem in calculations. Just carry it along like you would any other constant, like pi, so there is nothing you will have to do out of the ordinary.

Third, in the realm of further hints for solving this problem. One further hint is that the combined area equation becomes a quadratic in "x" if you use "x" for the circumference of the circle and "L - x" as the perimeter of the square. Another hint is that the coefficient for the x^2 term will be positive, so the graph is that of a parabola opening up. Another hint is that, because the area is of a parabola, you don't really have to use calculus to solve this problem, although this particular problem is a little easier solved using calculus than by simply completing the square. If you are really good at basic algebra, you might find just completing the square easier. Another hint is that if you use use calculus, you don't have to get the second derivative (unless you want to) if you are really sharp on properties of parabolas opening upward (or downward, for that matter). Last hint, and I really shouldn't give any more because you will otherwise have very little to do, is that BOTH parts of this problem make total sense. What I mean is that, at first, after solving the quadratic and getting either max or min, you might think that there ther is only a max or only a min. Keep in mind that we are not talking about unbounded "x". "x" is limited to working with the length of the wire which is L, so you are talking about getting "x" max and min in an INTERVAL which is a bit different from just looking at a parabola with x going off to +-infinity. There, I have now set up the problem, given you conversions for radius and circumference, gotten you back on track with the perimeter of the square, and given you lots of hints. That's about 80-90% done. The rest is up to you.