# Homework Help: Optimization Problem

1. Jul 21, 2012

### Plutonium88

1. The problem statement, all variables and given/known data
If the area of the window is 3m² what are the dimensions of a rectangle and (isosceles)triangle that will minimize the perimeter.
http://s12.postimage.org/r1czis8jh/optimize.png --DIAGRAM

2. Relevant equations

Let AreaTotal=3
Let Areatotal=Atriangle+Arectangle

Arectangle=lxw
Atriangle =bxh/2

3. The attempt at a solution

so because the triangle is isosceles the base and height are = x.

so Atriangle = x*x/2=x²
Arectangle= l*w=xy

Atotal = Atriangle+Arectangle
3=x²/2 + xy
y=(6-x²)/2x(1)-- y in terms of x

Prectangle= 2y+x
= 2(6-x²)/2x +x
=(6-x²)/x +x
=(6-x²+x²)/x
=6/x
it is only 2y+x because the perimieter of the window does not include the 4 rectangle side.

Ptriangle= x+x+√(2x²)
=2x+√(2x²)
The perimeter of the triangle includes opposite adjacent and hypotanuse sides.

Perimeter Total=Ptriangle + Prectangle.
Ptotal=6/x + 2x+√(2x²)

Ptotal=¸(6+2x²+x²*√(2))/x

Domain: XER, X Cannot Be Zero.

Now before i go any further, i was wondering if i could get some help ion figuring out if the equation i derived would be correct... Because from here, i would Take the derivative and determine critical points and a minimum value and then solves for the sides.. Anyway i appreciate any help :D

Last edited: Jul 21, 2012
2. Jul 21, 2012

### LCKurtz

But you don't include the whole perimeter of the triangle in the exterior perimeter either so the 2 highlighted below should be 1.
And writing $x\sqrt 2$ as $\sqrt{2x^2}$ will only complicate the derivative later.

3. Jul 21, 2012

### Plutonium88

Ah yea youre right, i dont know why i was including the side between the rectangle and triangle.. And okay, ill keep it as √(2x²), thank you for the tip.

im gonna rewrite it again, do you mind checking again?

4. Jul 21, 2012

### Plutonium88

So...

Ptriangle = x+2√x²
Prectangle= 2y+x=6/x

Ptotal = ptriangle+prectangle
ptotal=6/x + x + √2x²
ptotal=( 6 + x² + x√2x²)/x

ptotal=( x(2x+√2x²-2x/√(2x²))-(6+x²+ x√2x²) ) / x²

ptotal=(2x² + x√2x²-((x²/√(2x²)) - 6 - x² - x√2x²)/x²
ptotal= (x² - x²/√(2x²) - 6)/x²
ptotal= (x²(√(2x²)) - 6(√(2x²)) -x²)/x²(√(2x²))
ptotal= x(x²√2 -6√2 - x)/x²(√(2x²))
ptotal=(x²√2 -6√2 - x)/x(√(2x²))

now i have a quadratic in the numerator..
p=0
0=x²√2 -x -6√2

Is this correct ?

5. Jul 21, 2012

### LCKurtz

One leg plus hypotenuse gives $x+x\sqrt 2$.

6. Jul 21, 2012

### Plutonium88

ptriangle= x+x√2
prectangle=x+2y
=2(6-x²)/2x+x
=6/x

Ptotal=ptriangle+prectangle
Ptotal= 6/x + x +x√2

ptotal= (6 + x² + x²√2)/x

ptotal= ( (0 +2x + 2x√2)(x) - (6 + x² + x²√2)(1) )/x²

ptotal= (2x² + 2x²√2 -6 - x² - x²√2)/x²
ptotal=(x² + x²√2 -6)/x²

7. Jul 21, 2012

### LCKurtz

That's better. So what do you get for the dimensions and the minimum perimeter?

8. Jul 21, 2012

### Plutonium88

okay so..
if ptotal=(x² + x²√2-6)/x²

ptotal= dne
x=0 (restricted by domain)

ptotal=0
0=x² + x²√2-6
6 = x²(1+√2)
x=√(6/(1+√2))
x= +/-1.576

I so plotted this on a line..

**i made a chart here but it didnt work.. i basically took the values of the slope before and after the point 1.58

so what i found was

+ -1.58 -
positive before, negative after, so therefore that was a maxiumum.
and then

- 1.58 +
negative before positive after, so i found that this point was a minimum, so it had to be the minimum value.
**i plugged in values before and after and determined the sign of the slope.**

i found that 1.58 was a minimum value so therefore it is a CP.

So if this is the minimum value.

y=(6-x²)/2x
plug inx=1.58
y=1.109
y=1.11

Hypotenuse=√2x² = 2.23

so therefore dimensions are

x=1.58 for the triangle and rectangle, y=1.11 for the rectangle,
and hypotenuse of the triangle is hyp= 2.23

9. Jul 21, 2012

### LCKurtz

Your numbers are correct, but you keep doing that. You have written $\sqrt 2x^2$ which is the same as $x^2\sqrt 2$ which is not what you mean. If you insist on calling $x\sqrt 2$ as $\sqrt{2x^2}$ you need parentheses around the $2x^2$ or the top bar on the square root sign. It's easier to not put the $x$ under the square root sign in the first place, and in this situation it is best to write the $x$ before the $\sqrt 2$ to avoid such misunderstandings.

10. Jul 21, 2012

### Plutonium88

im really sorry about that that you for reminding me again, i need to watch that or people arent gonna understand me :(. Thank you very much for your help i really appreciate it :D.