# A rectangle has its two lower corners on the xx-axis and its two upper corners on the parabola y=6−x^2. What are the dimensions of such a rectangle with the greatest possible area?Width=Height=

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A rectangle has its two lower corners on the xx-axis and its two upper corners on the parabola y=6−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width=

Height=

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Step 1

Please see the white board. The blue line is the parabola. Rectangle has been shown in the green line. The dimensions have been represented by the red line help_outlineImage TranscriptioncloseHeight y 6-x Width 2x (-x, 0)! (x, 0) -2 -4 -6 -8 -10 -12 en st fullscreen
Step 2

Area of the rectangle, A = Width x height = 2xy = 2x(6 – x2) = 12x – 2x3

Step 3

For A to be maximum, the first derivative should be zero ad the second derivative at that point must be negative.

Hence, dA / dx = 12 - 6x2 = 0

Hence, x= ± √2

Since, x being a dimension, cn't be negative, hence, x = -√2 is rejeted.

d2A / dx2 = -12x < 0 for any or all the positive values of x

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