# Math Help - Optimization help needed..

1. I got a ' (x) =
-4x^2 +2 - 1.41 -x(sqroot2)=0

2. what happens when I multiply
2(1-x^2)^0.5 -(x)(sqroot2) all by (1-x^2)^0.5/(1-x^2)^0.5 ??

I think heres where i went wrong

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

$8x^4-7x^2+1=0$

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 $8x^2-7x+1=0$

Don't struggle with it.
If you post your calculations, they can be checked.

4. Originally Posted by wolfhound
what happens when I multiply
2(1-x^2)^0.5 -(x)(sqroot2) all by (1-x^2)^0.5/(1-x^2)^0.5 ??

I think heres where i went wrong
From $2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}-\sqrt{2}=0$

we multiply by $\sqrt{1-x^2}$ as the RHS is zero.

$2(1-x^2)-2x^2-\sqrt{2}\sqrt{1-x^2}=0$

$2-2x^2-2x^2=\sqrt{2-2x^2}$

$2-4x^2=\sqrt{2-2x^2}$

$(2-4x^2)(2-4x^2)=2-2x^2$

$4-8x^2-8x^2+16x^2=4-16x^2+16x^4=2-2x^2$

$16x^4-14x^2+2=0$

$8x^4-7x^2+1=0$

One of the positive solutions of this gives the maximum x

5. You can then try completing the square to solve the resulting equation.

$8x^4-7x^2+1=0$

$8x^4-7x^2=$

We first find "q".

$(\sqrt{8}x^2-q)(\sqrt{8}x^2-q)=8x^4-q\sqrt{8}x^2-q\sqrt{8}x^2+q^2=8x^4-2\sqrt{8}qx^2+q^2$

Therefore

$7=2\sqrt{8}q$

$q=\frac{7}{2\sqrt{8}},\ q^2=\frac{49}{4(8)}=\frac{49}{32}$

$q=\frac{7}{\sqrt{32}}$

Hence,

$8x^4-7x^2+\frac{49}{32}$ is a square

$8x^4-7x^2+\frac{32}{32}+\frac{17}{32}=\frac{17}{32}$

$(\sqrt{8}x^2-\frac{7}{\sqrt{32}})^2=\frac{17}{32}$

$\sqrt{8}x^2-\frac{7}{\sqrt{32}}=\pm\frac{\sqrt{17}}{\sqrt{32}}$

$\sqrt{8}x^2=\frac{7+\sqrt{17}}{\sqrt{32}},\ \frac{7-\sqrt{17}}{\sqrt{32}}$

$x^2=\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}},\ \frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}$

$x=\sqrt{\frac{7-\sqrt{17}}{\sqrt{32}\sqrt{8}}},\ \sqrt{\frac{7+\sqrt{17}}{\sqrt{32}\sqrt{8}}}$

One of these is the correct x maximum.

6. Here's a sketch of solving for the derivative = 0

7. 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

8. 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

$w=\sqrt{1-x^2}-\frac{1}{\sqrt{2}}$

$Area=2xw=2x[\sqrt{1-x^2}-\frac{1}{\sqrt{2}}]=2x\sqrt{1-x^2}-\frac{2x}{\sqrt{2}}=2x\sqrt{1-x^2}-\sqrt{2}x$

$\frac{d}{dx}A=2x(0.5[1-x^2]^{-0.5}(-2x))+2\sqrt{1-x^2}-\sqrt{2}=\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}-\sqrt{2}$

This is zero for maximum area, so the value of x that makes this zero must be found.

9. hello, sorry file was too big to upload here,
yes that what I got on the last line
-2x^2 +2(1_x^2) -sqr2=0
so -4x^2 +2 - 1.41 =0
-4x^2=0.59
x^2=-0.59/-4
sqr => x=0.383

a(0.383)= 0.166 ft^2

10. Ok,

you neglected to multiply $\sqrt{2}$ by $\sqrt{1-x^2}$.

you must multiply all 3 terms of the equation by $\sqrt{1-x^2}$

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!

11. Originally Posted by Archie Meade
Ok,

you neglected to multiply $\sqrt{2}$ by $\sqrt{1-x^2}$.

you must multiply all 3 terms of the equation by $\sqrt{1-x^2}$

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!
Hello again archie, ok so now I have sorted out this bit above, and worked everything down to
8x^4 - 7x^2 + 1 =0
so x^2 can be y
8y^2 -7y + 1 =0
.... y = 1/8 or y =1 so x = sqr 1/8 or x = sqr 1?? Help please?

12. Originally Posted by wolfhound
Hello again archie, ok so now I have sorted out this bit above, and worked everything down to
8x^4 - 7x^2 + 1 =0
so x^2 can be y
8y^2 -7y + 1 =0
.... y = 1/8 or y =1 so x = sqr 1/8 or x = sqr 1?? Help please?
that's a lot faster than what i did!

However, you can't factorise it to $(8y-1)(y-1)$

as it will give you a -9y term in the middle.

Instead, as $8y^2-7y+1$ doesn't factorise conveniently, use

For $ay^2+by+c=0,\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence $y=\frac{7\pm\sqrt{49-32}}{16}$

$x=\sqrt{y}=\sqrt{\frac{7+\sqrt{17}}{16}},\ \sqrt{\frac{7-\sqrt{17}}{16}}$

One of these is the correct x max

13. Originally Posted by Archie Meade
that's a lot faster than what i did!

However, you can't factorise it to $(8y-1)(y-1)$

as it will give you a -9y term in the middle.

Instead, as $8y^2-7y+1$ doesn't factorise conveniently, use

For $ay^2+by+c=0,\ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Hence $y=\frac{7\pm\sqrt{49-32}}{16}$

$x=\sqrt{y}=\sqrt{\frac{7+\sqrt{17}}{16}},\ \sqrt{\frac{7-\sqrt{17}}{16}}$

One of these is the correct x max
Hello archie, thanks for all your help, its such a relief to get it finished, I would have been lost without your help, I enjoyed it

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