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  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?


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up vote 1 down vote accepted

For 1, you are exactly right.

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.

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Thanks for your reply! I get 1 and 2 now but Im having troubles with the third question. Could you be kind enough and elaborate upon the question even further. It be great if you could solve it step by step :). Also By the sum of squares of the numbers do they means just the sum of the numbers which are squared? Thanks again! – John Aug 20 '11 at 22:49
@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
@John: As to whether you can "cheat this questions(s)" by taking as known a certain (true) property of the square, that unfortunately depends on whether the person grading agrees. (S)he might insist that you prove that the square indeed has that property, and then we are back at the original problem. – André Nicolas Aug 21 '11 at 3:39

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