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25.02.2020 •
Mathematics
(1 point) A rectangle is inscribed with its base on the xx-axis and its upper corners on the parabola y=11−x2y=11−x2. What are the dimensions of such a rectangle with the greatest possible area?
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Ответ:
the rectangle has dimensions of length= 2√(11/3) and height=22/3
Step-by-step explanation:
since the parabola
y= 11-x²
is symmetric with respect to the y-axis ( meaning that y(x)=y(-x) );
area of rectangle with base from -x to x and height y = 2 *area of rectangle with base from 0 to x and height y
A=2*x*y
replacing y
A=2*x*( 11-x²) = 22*x - 2*x³
the maximum value can be found when the derivative of the area with respect to x is 0 , thus
dA/dx= 22 - 6*x² = 0
22 - 6*x² = 0
x max = √(22/6)
y max = 11-xmax² = 11-[√(22/6)]²= 11 - 22/6 = 44/6
then the rectangle with maximum area has dimensions of
length = 2*x max= 2√(22/6) and height=y max=44/6
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