Show that the rectangle with the largest area that is inscribed within a circle of radius r is a square. Find the dimensions and the area of the inscribed square.
My respect goes to those who know how to tackle this one.
Let the base of the rectangle be x, let its height be y units.
the diagonal would be the diameter of the circle and it length is 2r.
so x^2 + y^2 = 4r^2
Area of rectangle = xy
= x√(4r^2 - x^2)
d(Area)/dx = ......
= (4r^2 - 2x^2)/√(4r^2 - x^2)
set this equal to zero for a maximum area and solve to get
x = r√2
put this back into x^2 + y^2 = 4r^2
to get y = r√2
so x=y, proving the rectange is a square
Thank you so much. I appreciate you taking the time to answer my question. = )
To prove that the rectangle with the largest area inscribed within a circle is a square, we can use mathematical reasoning and principles.
Let's consider a circle with radius r, and inscribe a rectangle within it as shown:
```
__________________
| |
| |
| |
| * |
| |
------------------
```
Let the rectangle have sides with lengths a and b. Since the rectangle is inscribed within the circle, the diagonal of the rectangle is equal to the diameter of the circle, which is 2r.
Using the Pythagorean theorem, we can express the relationship between the sides of the rectangle and the diagonal:
a^2 + b^2 = (2r)^2
a^2 + b^2 = 4r^2
To find the dimensions of the rectangle with the maximum area, we can solve for one variable in terms of the other. Let's solve for a in terms of b:
a^2 = 4r^2 - b^2
a = sqrt(4r^2 - b^2)
Now, we can find the area of the rectangle in terms of one variable:
Area = a * b = b * sqrt(4r^2 - b^2)
To find the maximum area, we can take the derivative of the area equation with respect to b and set it to zero:
d(Area)/db = sqrt(4r^2 - b^2) - (b^2 / sqrt(4r^2 - b^2)) = 0
Simplifying this equation, we can get:
sqrt(4r^2 - b^2) = b^2 / sqrt(4r^2 - b^2)
Squaring both sides of the equation, we get:
4r^2 - b^2 = b^4 / (4r^2 - b^2)
Cross-multiplying, we get:
(4r^2 - b^2)^2 = b^4
Expanding and simplifying this equation, we find:
16r^4 - 8r^2 b^2 + b^4 = b^4
16r^4 - 8r^2 b^2 = 0
Dividing both sides of the equation by r^2, we get:
16r^2 - 8b^2 = 0
8r^2 = b^2
Taking the square root of both sides, we find:
sqrt(8r^2) = sqrt(b^2)
sqrt(8) * r = b
So, we have found that b = sqrt(8) * r.
Now, substituting the value of b back into the equation for a, we find:
a = sqrt(4r^2 - (sqrt(8) * r)^2)
a = sqrt(4r^2 - 8r^2)
a = sqrt(-4r^2)
a = 0 (since the area cannot be negative)
From this, we can conclude that a = 0, which means the rectangle is degenerate and reduces to a line segment.
Therefore, the maximum area of the inscribed rectangle occurs when the rectangle is a square, with sides of length 2r (since b = sqrt(8) * r).
The dimensions of the inscribed square are 2r × 2r and the area is (2r)² = 4r².