The answer would be line segment.
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Answer:
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola
y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?
Width =
Height =
Width =√10 and Height ![= \frac{10}{4}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B10%7D%7B4%7D)
Step-by-step explanation:
Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)
are (h,k) and (-h,k).
Hence, the area of the rectangle will be (h + h) × k
Therefore, A = h²k ..... (2).
Now, from equation (1) we can write k = 5 - h² ....... (3)
So, from equation (2), we can write
![A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}](https://tex.z-dn.net/?f=A%20%3Dh%5E%7B2%7D%20%5B5-h%5E%7B2%7D%20%5D%3D5h%5E%7B2%7D%20-h%5E%7B4%7D)
For, A to be greatest ,
![\frac{dA}{dh} =0 = 10h-4h^{3}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdh%7D%20%3D0%20%3D%2010h-4h%5E%7B3%7D)
⇒ ![h[10-4h^{2} ]=0](https://tex.z-dn.net/?f=h%5B10-4h%5E%7B2%7D%20%5D%3D0)
⇒ ![h^{2} =\frac{10}{4} {Since, h≠ 0}](https://tex.z-dn.net/?f=h%5E%7B2%7D%20%3D%5Cfrac%7B10%7D%7B4%7D%20%7BSince%2C%20h%E2%89%A0%200%7D)
⇒ ![h = ±\frac{\sqrt{10} }{2}](https://tex.z-dn.net/?f=h%20%3D%20%C2%B1%5Cfrac%7B%5Csqrt%7B10%7D%20%7D%7B2%7D)
Therefore, from equation (3), k = 5 - h²
⇒ ![k=5-\frac{10}{4} =\frac{10}{4}](https://tex.z-dn.net/?f=k%3D5-%5Cfrac%7B10%7D%7B4%7D%20%3D%5Cfrac%7B10%7D%7B4%7D)
Hence,
Width = 2h =√10 and
Height = ![k =\frac{10}{4}.](https://tex.z-dn.net/?f=k%20%3D%5Cfrac%7B10%7D%7B4%7D.)