I'm in grade 9 and very soon, I'm going to have my final exam in math. Now, I'm pretty decent with almost everything I learned in math, except the optimization of shapes (especially rectangles.) My question is: How do you optimize rectangles, so that you can get the maximum area, or the minimum area?
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2Maximum area - make it a square. Minimum area - make it a line (I could prove it to you with derivatives, but I assume they don't ask you to do that in grade 9). By the way, you probably forgot to mention that the circumference of the rectangle should remain the same. – barak manos Jan 21 '15 at 12:52
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1Normally in problems like this you express the quantity you want to minimize (or maximize) as a quadratic function. You know how to minimize (or maximize) a quadratic function by completing the square. It might help if you gave a specific example of a problem you don't know how to solve. – littleO Jan 21 '15 at 12:59
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make some rectangles while keeping the area fixed. that means you need to one sheet of rectangular paper and keep reducing the length and increasing the width without using any extra paper. – abel Jan 21 '15 at 13:47
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The area of a rectangle with sides $a$ and $b$ is $$ A = a b $$
The perimeter of a rectangle with sides $a$ and $b$ is $$ P = 2 (a + b) $$
Assuming we have a given (fixed) perimeter $P$, how should we choose $a$ and $b$ to maximize the area $A$?
If the perimeter is a constant, we can use the relationship above to express one side by a term of the other side and the perimeter constant: $$ P = 2(a+b) \Rightarrow b = \frac{P}{2} - a $$
This we can insert into the formula for the area: $$ A = a b = a\left(\frac{P}{2} - a\right) = a \frac{P}{2} - a^2 $$ to express the are in terms of side $a$ and perimeter $P$.
This formula can be used to find the value for $a$ which maximizes $A(a)$. Here is a plot for $P=4$.
It suggests that $a = 1$ is the best choice. In this case $b=4/2 - 1 = 1$ as well, so this the case for a square.
mvw
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