Question

need help on the attachment need help on the attachment @Mathematics

Answer 1

\[y={6x+8\over x^2+1}\]because it is a rectangle\[A(x)=xy={6x^2+8x \over x^2+1}\]to find max/min we need the derivative of this:\[A'(x)={(12x+8)(x^2+1)-2x(6x^2+8x)\over (x^2+1)^2}={-4(2x^2-3x-2)\over (x^2+1)^2}=0\]so we need the roots of \[2x^2-3x-2=0x={3 \pm \sqrt{(-3)^2-4(2)(-2)} \over 2(2)}={3\pm5\over4}=\left\{ -1/2,2 \right\}\]normally for an area only the positive root would make sense, but because this rectangle is based on a graph that can have a negative x-coordinate as one side, so we have to check\[A(2)=8\]\[\left| A(-1/2) \right|=\left| -2 \right|=2\]so it looks like 8 units is our maximum.

Answer 2

Thanks!

Answer 3

Also why is the numerator 6x^2+8x

Answer 4

also did you plug -1/2 into the original equation to get 2

Answer 5

The area of the square A(x) is xy
multiply the expression you have for y by x and you will get that numerator.

All critical numbers were plugged into A(x)=xy, so no I did not plug anything into the original equation because the original equation is just for the y-component, not the area of the rectangle