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, rounded to the nearest 3 decimals?
2 Answers
Length:
Height:
The maximum area is approximately
Explanation:
Let the two lower vertex of the rectangle be
The area is therefore:
#A = 2xy#
But we want to eliminate one of these variables. We see that for the context of this problem,
#A = 2x(5 - x^2)#
#A = -2x^3 + 10x#
Now we differentiate to find the critical point(s).
#A' = -6x^2 + 10#
This will have critical point(s) when it equals
#0 = -6x^2 + 10#
#0 = -2(3x^2 - 5)#
#3x^2 = 5#
#x = +-sqrt(5/3)#
Therefore, the length will be
Hopefully this helps!
Maximum area of
Explanation:
Let
Let us set up the following variables:
# { (alpha,"semi-width of rectangle"), (beta,"height of rectangle") (A,"Area of rectangle") :} #
Our aim is to find
As
# beta = 5-alpha^2 \ \ \ \ \ ..... [1]#
And the total Area is that of a rectangle of width
# A = 2 alpha beta #
# \ \ \ = 2 alpha (5-alpha^2) \ \ \ \ \# (from [1] )
# \ \ \ = 10alpha-2alpha^3 #
We now have the Area,
# (dA)/(d alpha) = 10-6alpha^2 #
At a critical point we have
# 10-6alpha^2 = 0 #
# :. alpha^2 = 5/3 => alpha = +-sqrt(5/3) #
We require that
With this value of
# beta = 5-alpha^2 #
# \ \ = 5-5/3 #
# \ \ = 10/3 ~~ 3.333333 #
And the corresponding area is:
# A = 2 alpha beta #
# \ \ \ = 2 sqrt(5/3) 10/3 ~~ 8.6066296 ...#
We can visually verify that this corresponds to a maximum by looking at the graph of
graph{10x-2x^3 [-2, 4, -5.5, 10]}
Which is consistent with a maximum of
Thus:
Maximum area of
Where the width is