Asked by jim
A rectangle has its base on the x-axis and its upper two vertices on the parabola
y=12−x2 .
What is the largest area that the rectangle can have
y=12−x2 .
What is the largest area that the rectangle can have
Answers
Answered by
Reiny
let the points of contact be (x,y) and (-x,y)
so the base is 2x and the height is y
but y=12-x^2
so the
Area = 2xy
= 2x(12-x^2)
= 24x - 2x^3
d(Area)/dx = 24 - 6x^2
= 0 for a max Area
solve 2x^3 - 24 = 0
sub that back into the Area = 2x(12-x^2)
so the base is 2x and the height is y
but y=12-x^2
so the
Area = 2xy
= 2x(12-x^2)
= 24x - 2x^3
d(Area)/dx = 24 - 6x^2
= 0 for a max Area
solve 2x^3 - 24 = 0
sub that back into the Area = 2x(12-x^2)
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