Answer :
If one of the vertices on the x-axis is (x, 0), then the other vertex on the same axis is (-x, 0), so that the rectangle has base 2x. The other two vertices on the parabola are the points (x, 6 - x²) and (-x, 6 - x²), so the height of the rectangle is 6 - x².
Then the area of the rectangle is given by the function
[tex]A(x)=2x(6-x^2)=12x-2x^3[/tex]
Compute the critical points of A:
[tex]A'(x)=12-6x^2=0\implies x=\pm\sqrt2[/tex]
So the maximum area is obtained when the vertices are the points (-√2, 0), (√2, 0), (√2, 4), and (-√2, 4). This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.
The maximum area is obtained when the vertices are the points
(-√2, 0), (√2, 0), (√2, 4), and (-√2, 4).
This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.
The area of a rectangle is expressed as shown:
A = xy
x is the length of the rectangle
y is the width
Given that y = 6 - x²
If its other two vertices above the x-axis and lying on the parabola, hence
the length of the rectangle will be 2x
The area of the rectangle will be A = 2x(6-x²)
If the dimension of the rectangle is at the maximum, hence dA/dx=0
A = 12x - 2x³
dA/dx = 12 - 6x²
0 = 12 - 6x²
6x² = 12
x² = 12/6
x² = 2
x = ±√2
Hence the maximum area is obtained when the vertices are the points
(-√2, 0), (√2, 0), (√2, 4), and (-√2, 4).
This rectangle has base 2√2 and height 4, giving a maximum area of 8√2.
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