The equation for the area of a rectangle is A = lw, where l is the length of one side and w is the width. In order to find the largest possible rectangle that can be inscribed in an ellipse x2 a2 + y2 b2 = 1, we must first find what values work with this equation. For instance, if we plug in 2a as our length and 2b as our width then A becomes 8ab; which does not work because it’s not equal to 1. If we plug in 3a as our length and 3b as their width then A becomes 12ab – which also doesn’t work because it’s not equal to 1. Finally, if we plug in 4a for our length and a for our width then A becomes 16a which does work because it’s equal to the area of an ellipse. So, we find that the largest rectangle that can be inscribed in this particular ellipse is one with dimensions x = -sqrt(b)and y= sqrt(-a).
In order to find what values would work with this equation, we first have to find out what works as lengths (x) and widths (y). For instance if we plug in two different numbers such as (x) being 12cm long by 20 cm wide for a rectangle then A becomes 12*20 = 240cm.
However, if we find the width to be sqrt(-a) instead of 20cm (sqrt(b)) as in example x above, then this would not work because it’s not equal to 240cm squared which is what the original equation tells us should happen.
This math problem can be solved with logarithms and exponents by first finding the length of one side of a square that is inscribed inside an ellipse. The area of a square will always be lw^ – or just wl^ when both lengths are equal such as they are here so all we have to do now is plug in our values into that equation.
The equation is x^a+y^b=lw which after plugging in our values becomes 12*20 = 320cm and we solve for A:A=(x)^(a)+((y) * 20)* ((b)) so with the values of a, b, l from above this would be 64 + (320 * (-12)).
We find that when both x and y are multiplied by -12 then they will cancel out leaving only their product as seen here to simplify things. So what’s left over is something squared plus something else times itself or just two squaring operations combined into one fractional operation; since there’s no division sign it means that ‘multiply’: square root
What we’re really after here are values for x and y which will cause this to be a maximum area rectangle, corresponding to an ellipse with the given equation (x^a+y^b=lw). This is simply accomplished by setting these variables so that they are equal in magnitude but opposite in sign:
A=-12 as seen before and b becomes 12 instead of -12–since negative numbers behave like their opposites when multiplied against themselves.Our found rectangle is now a 12 by 24 rectangle with an area of 288: the maximal possible value for such a shape inscribed in this ellipse.
The answer to our problem, then, is that the largest area rectangle that can be inscribed in this circle has dimensions 12x24and an area of 288 square units.
Notice how we were able to find both the width and the length of our rectangle, which is not always possible when we try to find areas.We can also find a maximal area rectangle by using coordinates -12i 12j as (xl yh) instead of (-12,-24). This will give us an area of 216 square units. The answer to our problem then becomes that the largest area rectangle that can be inscribed in this circle has dimensions 12×24 and an area of 288 or 216 square units.
