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Algebra 19 Online
OpenStudy (trojanpoem):

Find the maximum value:

OpenStudy (trojanpoem):

|dw:1428098264490:dw|

OpenStudy (trojanpoem):

I manged to get the minimum value which is x = 2 now I am stuck with the maximum value f(x) = 3/2 * x^2 -6x + 72

OpenStudy (mathmath333):

is \(\square ABCD\) a rectangle, parallelogram \(\cdots\) ?

OpenStudy (trojanpoem):

neither rectangle nor parallelogram. It's a square of side length = 12 cm.

OpenStudy (mathmath333):

and u want maximum value of which thing ?

OpenStudy (trojanpoem):

We went the maximum and minimum values for the area of the triangle inscribed inside the square.

OpenStudy (trojanpoem):

I supposed that It would be the area of square - 3 triangles and got the equation above.

OpenStudy (mathmath333):

and \(AE=x\) and \(FB=3x\) that's the only information given

OpenStudy (trojanpoem):

The side length of the square is 12 cm

OpenStudy (trojanpoem):

Oo , but the book answer keys says that the minimum value is at 2 when the area = 66 and the maximum at two points and the area is 72.

OpenStudy (trojanpoem):

|dw:1428101114405:dw| area = 144 - 6x - 6(12 - 3x) - 0.5 * 3x * (12-x) area = 3/2 x^2 - 6x + 72 da/dx = 3x -6 when da/dx = 0 x = 2 area= 66 cm^2

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