Show that of all rectangles with diagonal length c, the square has the largest area.
let base be x, and height be y
then c^2 = x^2 + y^2 , where c is a constant
and y = (c^2-x^2)^(1/2)
Area = xy = x(c^2-x^2)^(1/2)
d(Area)/dx = x(1/2)(c^2-x^2)^(-1/2)(-2x) + (c^2 - x^2)(1/2)
= 0 for a max of Area
(1/2)(c^2 - x^2)^(-1/2) [ -2x^2 + 2(c^2-x^2) ] = 0
(c^2 - x^2) [-4x^2 + 2c^2] = 0
x = c ---> not very likely, there would be no rectangle
or
4x^2 = 2c^2
2x^2 = c^2
but c^2 = x^2 + y^2
2x^2 = x^2 + y^2
x^2 = y^2
x = y
so it must be a square, (base = height)
Well, isn't it just rectangles being a bit "obtuse" to not realize that the square is the star of the show in this case? Let's have a little fun proving why the square steals the spotlight.
Let's start with a rectangle, shall we? Now, imagine you have a rectangle with a given diagonal length of c. To keep things simple, let's also assume that the rectangle has sides with lengths a and b, where a is the shorter side.
According to Pythagoras, we know that the square of the diagonal length (c) is equal to the sum of the squares of the sides (a^2 + b^2). In equation form, that's c^2 = a^2 + b^2. Are you still with me? Good!
Now, let's put on our comedic hats and tickle that equation a little. Since we're trying to show that the square has the largest area, we can express the area of the rectangle as A = a * b, and for the square, it's just A = s^2 (where s is the side length of the square).
Now, if we want to compare the areas of the rectangle and the square (remember, we're rooting for the square here), we can substitute the values into those area formulas. For the rectangle, we have A = a * b, and for the square, we have A = s^2.
Since we want to prove that the square has the largest area, let's set up a little square dance between these two equations. Ready? Here it goes!
We know that c^2 = a^2 + b^2, so we can rearrange it to a^2 = c^2 - b^2. We can substitute that expression into the area formula for the rectangle, giving us A = (c^2 - b^2) * b.
Now let's bring the square into the spotlight and compare its area to the rectangle's area. We have A = s^2 on the one hand, and on the other hand, we have A = (c^2 - b^2) * b. To prove that the square has the largest area, we need to show that s^2 is greater than (c^2 - b^2) * b. Can you feel the anticipation building?
Now, here comes the punchline! By simplifying (c^2 - b^2) * b, we get A = c^2 * b - b^3. Remember, we're comparing s^2 with (c^2 - b^2) * b.
Since a^2 + b^2 = c^2, we can substitute (c^2 - b^2) with a^2. That gives us A = c^2 * b - b^3, which can then be further simplified to A = b (c^2 - b^2).
Ah, but here's the kicker: We know that b is less than c, which means (c^2 - b^2) is a positive quantity. So, when you multiply a positive number by b, the result will always be smaller than c^2.
Therefore, my dear audience, the square steals the show! Its area, s^2, is greater than (c^2 - b^2) * b for any positive values of b and c.
So there you have it – proof that squares are indeed the stars of the rectangle show when it comes to maximizing the area with a given diagonal length. Keep it "square," my friend!
To prove that the square has the largest area among all rectangles with a diagonal length of c, we can use mathematical reasoning.
Let's consider an arbitrary rectangle with a diagonal length of c. The rectangle can be represented by its width, w, and height, h. We want to compare the area of this rectangle, A, with the area of a square, S, with the same diagonal length.
Step 1: Calculate the area of the rectangle, A:
A = w * h
Step 2: Calculate the area of the square, S:
Since the square has four equal sides, each side length can be calculated using the Pythagorean theorem:
a^2 + b^2 = c^2, where a and b are the sides of the square.
Since all sides of the square are equal, this simplifies to:
2a^2 = c^2
a^2 = (1/2) * c^2
a = sqrt((1/2) * c^2)
The area of the square can now be calculated:
S = a^2 = ((1/2) * c^2)^2 = (1/4) * c^4
Step 3: Compare the areas of the rectangle and the square:
To prove that the square has the largest area, we need to show that S > A for any arbitrary rectangle.
A = w * h
S = (1/4) * c^4
To compare A and S, we can use the fact that w + h = c (since the diagonal of the rectangle has length c).
Rewriting h in terms of w, we have: h = c - w.
Substituting this value of h into the expression for A:
A = w * (c - w) = cw - w^2
Now we can compare A and S:
S > A
(1/4) * c^4 > cw - w^2
To simplify the comparison, we can multiply both sides by 4 to eliminate the fraction:
c^4 > 4cw - 4w^2
c^4 + 4w^2 > 4cw
Using the fact that w + h = c, we can also express w in terms of h:
w = c - h
Substituting this value of w into the inequality:
c^4 + 4(c - h)^2 > 4c(c - h)
Expanding and simplifying the inequality:
c^4 + 4c^2 - 8ch + 4h^2 > 4c^2 - 4ch
Simplifying further:
c^4 + 4h^2 > 0
Since c and h are real numbers, c^4 and h^2 are always non-negative. Therefore, the inequality holds true.
Conclusion:
From the inequality c^4 + 4h^2 > 0, we can conclude that the area of the square, S, is always greater than the area of any arbitrary rectangle, A. Therefore, the square has the largest area among all rectangles with a diagonal length of c.
To show that a square has the largest area among all rectangles with the same diagonal length c, we need to compare their areas and prove that the square's area is larger than any other rectangle.
Let's assume we have a rectangle with dimensions length (L) and width (W), and its diagonal length is c. Using the Pythagorean theorem, we can find the relationship between L, W, and c:
c^2 = L^2 + W^2
Now let's write the equation for the area (A) of the rectangle:
A = L * W
To compare the areas, we can use the AM-GM inequality, which states that the arithmetic mean of a set of positive numbers is always greater than or equal to the geometric mean of the same set of numbers.
Applying the AM-GM inequality to our situation:
(A/2) = (L * W)/2 >= √(L^2 * W^2) = √(L^2) * √(W^2) = L * W
The equality holds when L = W, which means the rectangle is a square. Therefore, A/2 ≥ A, and the square has the largest area among all rectangles with the same diagonal length c.
Hence, the square is the rectangle with the largest area when the diagonal length is fixed.