# Maximum of functions word problem

1. Find the dimensions of the rectangular field of maximum area which can be enclosed with 144 meters of fence. Am I right to say that you can cheat this questions by knowing that squares have the largest surface area, thus you can just divide the perimeter into 4? You could solve for the Length and create a function: A = (72 - w) * w. Then You could find the the maximum of said function and look at the x value (width) and solve L = (144 - 2x)/2. Is the second set of calculations correct and applicable to harder questions? The correct answer is 36*36 right?

2. Divide 56 into two parts whose product is a maximum. I know its 28, but how do I write out the correct set of calculation, for when the questions get harder?

3. Here is a harder question I found: Find two numbers whos sum is 18 if the sum of the squares of the numbers is a minimum. What do they mean by sum of the squares of the numbers? How do I approach this question?

Thanks!

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For 2, it is the same as 1 with a perimeter of $2*56=112$. The product is $p=x(56-x)$ and you can find the maximum the same way. You will get $28$ as you say.
For 3, if one number is $x$, the other is $18-x$. Then you are asked to minimize $x^2+(18-x)^2$. The same technique you have used in 1 will work again.
@John: The sum is $s=x^2+(18-x)^2=324-36x+2x^2$ Taking the derivative and setting to zero we have $0=-36+4x$, which gives $x=9$. You can check this by making a spreadsheet and checking the value of $s$ as $x$ goes from $0$ to $18$. – Ross Millikan Aug 20 '11 at 22:54
@John: You can minimize $324-36x+2x^2$ by minimizing $-36x+2x^2$, that is, by maximizing $2(x)(18-x)$. So we want to maximize $x(18-x)$, and we are back to the previous problems. However, one can't avoid the calculus forever! – André Nicolas Aug 21 '11 at 3:04