<p>You have a 12-inch wire. You cut it into two pieces. You bend one into the shape of a circle and the other into the shape of a square. How should the pieces be cut so that the area is a) a maximum? b) a minimum?</p>
<p>to maximize, se the derivative equal to zero. first, you need to find the function to maximize, in this case area. for the circle, A=pi<em>r^2 for the square, A=s^2. in maximization/minimization problems, the most important aspect is finding one more way to relate the variables. since the only other piece of information given is that the wire is 12 inches long, you know that you need to use that. the total amount of wire used will be 12, so the perimeters of the 2 shapes will add up to 12. therefore 12=2</em>pi<em>r+4s. the total area is A=pi</em>r^2+s^2 so no you have to eliminate one variable because you can’t derive with respect to 2 different variables at once. 3-pi/2<em>r=s and substituting back into the area function A=pi</em>r^2+ (3-pi/2<em>r)^2. NOW you derive with respect to r and you get A’=2</em>pi<em>r+ (3-pi/2</em>r)<em>(-pi/2) (used chain rule for the second part). setting A’=0 you get 0=2pi</em>r-3pi/2+(pi/2)^2<em>r which you can plug into a calculator. in that case r=.53855 s=2.154 (plugged back into the 12= formula) and total area is 5.55. *</em><em>BUT you also have to check the endpoints of the function, in this case where one shape has a side length or radius of 0. so the area when the wire only makes up the circle is c=12 so A=1.9098, but when the wire only makes up the square the area is 4</em>4=16. therefore the wire should just be used to make a square in order to maximize area and just a circle in order to minimize area.</p>
<p>Maximun: make a circle
Mininum: mae a square</p>
<p>This is just a geometry problem, no calc involved.</p>
<p>actually maximization and minimization (otherwise known as optimization) is on the calc AB/BC topic outlines and yes, this problem is calculus. you could guess and get lucky because it’s the endpoints in this case, but a geometric proof of this problem would be very difficult.</p>
<p>lol just want to point out that “but when the wire only makes up the square the area is 4<em>4=16. therefore the wire should just be used to make a square in order to maximize area and just a circle in order to minimize area” Wouldn’t a 4</em>4 make the wire 16 inches long.</p>
<p>fair enough, 3x3 square with an area of 9. I got a 5 on calc AB but only 720 on the SAT math 
plus most of my mistakes this year in calc were number mistakes rather than calculus mistakes.</p>
<p>2<em>Pi</em>r=12 => r=6/Pi. The area of circle = 36/Pi. The area of square = 9, which is less than 36/Pi. Thus when the wire form a circle, its area is greatest.</p>