Why is a 35 x 35 foot room bigger by 25 square feet than a 30 x 40 foot room?

Answer 1

Algebra is a good tool for showing this. We know that the area of any square or rectangular region of length l and width w is given by: A = l w Now, take a value called "x" and let it equal 35. Then we can say that the area of the rectangle, 35 by 35, would be x times x which is x2 . Now consider the area of the 30 by 40 rectangle. Instead of x times x, it would be: (x - 5)(x + 5) If we multiply those out, it gives us a "difference of squares", which comes to:

x2 - 25

and is clearly 25 less than x2 .

This will hold true no matter what the difference is, giving us an algebraic demonstration why the 35 by 35 has a 25 square feet greater area.

This can also be shown using a little bit of calculus:

First, consider the perimeter of the room to to be a number that must be kept the same. In that case we have 140 feet of perimeter used by both of those rooms. We know that the area is equal to the length multiplied by the width:

a = l * w

We also know that, given our limited total perimeter, the length and width have a direct relationship as well:

P = 2l + 2w

∴ 2l = P - 2w

∴ l = P/2 - w

In this case, P is a given, so we can say:

l = 140/2 - w

∴ l = 70 - w

We can now take that value of l, and replace it in our equation for area:

a = (70 - w) * w

∴ a = 70w - w²

This equation gives us the relationship between the width of the room and it's total area, given that it's perimeter is 140 feet. At this point, you can plug values in to see which ones give you the most area. For example:

if w = 30, then a = 70 * 30 - 30 * 30 = 40 * 30 = 1200

No matter what values you plug in, you'll never get a better number than 1225, which is what you get when w = 35. But you can also prove mathematically that there is no better number with a little bit of calculus.

If you look at our equation for the area, a = 70w - w², you can see that it defines an upside down parabola. This means that the most area can be found when that parabola is at it's peak.

Note that the slope of the parabola's tangent will be 0 at it's peak. This means that we can find that ideal point by taking the derivative of the equation (which gives us the slope of it's tangent) with respect to it's width, and solving it for 0.

if a = 70w - w²

then a' = 70 - 2w

We then solve for a' = 0:

0 = 70 - 2w

∴w = 70 / 2

∴w = 35