I got a ' (x) =
-4x^2 +2 - 1.41 -x(sqroot2)=0
The negative answers are just algebraic answers from the graph,
but we must work with positive values as we are dealing with lengths of the rectangle sides.
After calculating the derivative, setting it to zero, rearranging the form i wrote, we end up with
$\displaystyle 8x^4-7x^2+1=0$
This has 4 solutions, two of them are negative though.
There is one correct positive answer corresponding to maximum x.
The other solution is extraneous and is caused by squaring terms to end up with $\displaystyle 8x^2-7x+1=0$
Don't struggle with it.
If you post your calculations, they can be checked.
I will add something later.
From $\displaystyle 2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}-\sqrt{2}=0$
we multiply by $\displaystyle \sqrt{1-x^2}$ as the RHS is zero.
$\displaystyle 2(1-x^2)-2x^2-\sqrt{2}\sqrt{1-x^2}=0$
$\displaystyle 2-2x^2-2x^2=\sqrt{2-2x^2}$
$\displaystyle 2-4x^2=\sqrt{2-2x^2}$
$\displaystyle (2-4x^2)(2-4x^2)=2-2x^2$
$\displaystyle 4-8x^2-8x^2+16x^2=4-16x^2+16x^4=2-2x^2$
$\displaystyle 16x^4-14x^2+2=0$
$\displaystyle 8x^4-7x^2+1=0$
One of the positive solutions of this gives the maximum x
You can then try completing the square to solve the resulting equation.
$\displaystyle 8x^4-7x^2+1=0$
$\displaystyle 8x^4-7x^2=$
We first find "q".
$\displaystyle (\sqrt{8}x^2-q)(\sqrt{8}x^2-q)=8x^4-q\sqrt{8}x^2-q\sqrt{8}x^2+q^2=8x^4-2\sqrt{8}qx^2+q^2$
Therefore
$\displaystyle 7=2\sqrt{8}q$
$\displaystyle q=\frac{7}{2\sqrt{8}},\ q^2=\frac{49}{4(8)}=\frac{49}{32}$
$\displaystyle q=\frac{7}{\sqrt{32}}$
Hence,
$\displaystyle 8x^4-7x^2+\frac{49}{32}$ is a square
$\displaystyle 8x^4-7x^2+\frac{32}{32}+\frac{17}{32}=\frac{17}{32}$
$\displaystyle (\sqrt{8}x^2-\frac{7}{\sqrt{32}})^2=\frac{17}{32}$
$\displaystyle \sqrt{8}x^2-\frac{7}{\sqrt{32}}=\pm\frac{\sqrt{17}}{\sqrt{32}}$
$\displaystyle \sqrt{8}x^2=\frac{7+\sqrt{17}}{\sqrt{32}},\ \frac{7-\sqrt{17}}{\sqrt{32}}$
$\displaystyle x^2=\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}},\ \frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}$
$\displaystyle x=\sqrt{\frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}},\ \sqrt{\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}}}$
One of these is the correct x maximum.
Hello archie
thanks again for your help and time,
tomorrow I will do it again and use your methods, because I have been doing it again, and this time I got 0.166 ft^2 absolute max
I dont know why I didnt get 0.168 but my way isnt the quadratic
please look at my pic View image: recprob1.jpg
thanks
Well done, wolfhound,
0.166 is close enough (it's within 2 thousandths of a square foot of 0.168).
The image is difficult to make out unfortunately...
Here are my calculations for the derivative
$\displaystyle w=\sqrt{1-x^2}-\frac{1}{\sqrt{2}}$
$\displaystyle Area=2xw=2x[\sqrt{1-x^2}-\frac{1}{\sqrt{2}}]=2x\sqrt{1-x^2}-\frac{2x}{\sqrt{2}}=2x\sqrt{1-x^2}-\sqrt{2}x$
$\displaystyle \frac{d}{dx}A=2x(0.5[1-x^2]^{-0.5}(-2x))+2\sqrt{1-x^2}-\sqrt{2}=\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}-\sqrt{2}$
This is zero for maximum area, so the value of x that makes this zero must be found.
Ok,
you neglected to multiply $\displaystyle \sqrt{2}$ by $\displaystyle \sqrt{1-x^2}$.
you must multiply all 3 terms of the equation by $\displaystyle \sqrt{1-x^2}$
It does involve much more calculations, unfortunately!
and the value for "x max" is a bit off.
it does give a very close answer, however.
Well, i hope the exercise was beneficial, wolfhound.
God loves a tryer!
that's a lot faster than what i did!
However, you can't factorise it to $\displaystyle (8y-1)(y-1)$
as it will give you a -9y term in the middle.
Instead, as $\displaystyle 8y^2-7y+1$ doesn't factorise conveniently, use
For $\displaystyle ay^2+by+c=0,\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Hence $\displaystyle y=\frac{7\pm\sqrt{49-32}}{16}$
$\displaystyle x=\sqrt{y}=\sqrt{\frac{7+\sqrt{17}}{16}},\ \sqrt{\frac{7-\sqrt{17}}{16}}$
One of these is the correct x max