Maximize area

1. May 6, 2009

joemama69

1. The problem statement, all variables and given/known data

a pentagon is formed by placing an isosceles triangle on a rectangle. If the pentagon has a fixed perimeter P, determine the lengths of the sides of the pentagon that maximize the area of the pentagon

2. Relevant equations

3. The attempt at a solution

not sure where to start since it is a irregular pentagon

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2. May 6, 2009

tiny-tim

What's the difficulty?

Call the lengths of the sides a b and c, then add the area of the rectangle to that of the triangle, and maximise.

3. May 6, 2009

FedEx

I think there is a problem. Cause we are dealing with three variables but we have just two equtions 1) 2a + 2b + c =P and 2) The one which tells about the total area. By normal convention we diff equation number 2. But we have three variables out of which we are able to eliminate just one.

Or else we could convert the equation number 2) in a quadratic of b. I am not totally sure whether that works or not

4. May 6, 2009

Dick

Why do you think there is a problem? The perimeter constraint eliminates one variable. Now you have an equation for the area in terms of two variables. You 'diff' equation 2 with respect to each one of the two remaining variables and set them equal to zero. That's two equations in two unknowns. Where is problem?

5. May 7, 2009

FedEx

Are you telling me to differentiate it partially?

Well course.

Well the thing is that we havent been taught with partial differentiation. Theres not much to it in undergraduate math... but still it does make a diff if you have learnt it formally

6. May 8, 2009

joemama69

heres what i got so far

A(1) = ab
A(2) = .5a$$\sqrt{c^2 - (a/2)^2}$$

Total A = A(1) + A(2) = ab + .5a$$\sqrt{c^2 - (a/2)^2}$$

P = a + 2b + 2c

"Why do you think there is a problem? The perimeter constraint eliminates one variable."

by that do mean solving for lets say a = P - 2b - 2c
and then i would plug it into the other equation

A = (P - 2b - 2c)b + .5(P - 2b - 2c) $$\sqrt{c^2 - (P - 2b - 2c) /2^2}$$

"Now you have an equation for the area in terms of two variables"

seems like theres three to me, P,b,c

"You 'diff' equation 2 with respect to each one of the two remaining variables and set them equal to zero."

which variables do i differentiate in terms of

7. May 8, 2009

Dick

P isn't a variable. It's 'fixed'. Your final answers for a, b and c will need to be given in terms of P.

8. May 8, 2009

FedEx

Precisely. So diff P w.r.t to any variable would give you zero

9. May 11, 2009

joemama69

ok

A = Pb - 2b2 - 2bc + (P/2 - b - c)(2c2 - P2/4 + Pc + Pb + b2 + 2bc)1/2

Ab = P - 4b + 2c + 1/2(p/2 - b - c)(P + 2b + 2c)(2c2 - p2/4 + Pc + Pb + b2 + 2bc)-1/2 - (2c2 - p2/4 + Pc + Pb + b2 + 2bc)1/2

Ab = P - 4b + 2c + (P2/8 - b2 - 2c2 - 3bc - Pc/2)/(2c2 - P2/4 + Pc + Pb + b2 + 2bc)1/2 - (2c2 - P2/4 + Pc + Pb + b2 + 2bc)1/2

So that is the Partial of A with respect to b. Based on what u told me i will also have to differentiate A with respect to c. I should set them both equal to 0, find all critical points, and then figure out which one is the maximum.

Have i done it right so far. Its quite messy, is there an easier way or do i just have to deal w/ it

10. May 11, 2009

Dick

I think it's somewhat messy however you do it. But doing it that way I definitely would have eliminated b instead of a. The easiest way is probably to put P in as a Lagrange multiplier, if you've learned that.

11. May 12, 2009

joemama69

Nope we never learned that. this is pretty messy, if you think its work it do u mind walkin me through it

BTW, i solved bor b instead, much simpler

A'(a,c) = P/2 - 2a - c + (2ac - a^2/2)/(4$$\sqrt{c^2 - a^2/4}$$) + $$\sqrt{c^2 -a^2/4}$$/2

12. May 12, 2009

Dick

I would start with the d/dc equation rather than the d/da. It's MUCH simpler. It's pretty easy to find the relation between a and c and eliminate another variable in d/da.

13. May 12, 2009

joemama69

I assume that means the partial of A with respect to c

Ac = -a + ac / (2$$\sqrt{c^2 - a^2/4}$$) = 0

a = ac / (2$$\sqrt{c^2 - a^2/4}$$) Square both sides to solve for a

a2 = a2c2 / (4c2 - a2)

4a2c2 - a4 = a2c2

a2 = 4c2 - c2 therefore a = c$$\sqrt{3}$$

0 = -c$$\sqrt{3}$$ + (c2$$\sqrt{3}$$) / (2(c2 - 3c2/4)1/2)

0 = -c$$\sqrt{3}$$ + (c2$$\sqrt{3}$$) / 2(c2/4)1/2

0 = -c$$\sqrt{3}$$ + (c2$$\sqrt{3}$$) / c

0 = -c$$\sqrt{3}$$ + c$$\sqrt{3}$$

something aint right. where i go wrong

14. May 12, 2009

Dick

I'll agree with a=sqrt(3)*c. Beyond that I think you should learn to check you own work. Would you try it again and if it doesn't work post clearly what you got for dA/da. I agree with parts of what you got, but I didn't get the same form for the whole thing. I think you are capable of finishing this on your own. That's a compliment.

15. May 13, 2009

joemama69

Yup got the same thing

Ok so

Ac = -a + (ac)/(2$$\sqrt{c^2 - (a^2)/4}$$) = 0, a = $$\sqrt{3}$$c

0 = -$$\sqrt{3}$$c + ($$\sqrt{3}$$c2)/(2$$\sqrt{c^2 - (3c^2)/4}$$ )

0 = -$$\sqrt{3}$$c + ($$\sqrt{3}$$c2)/(2$$\sqrt{(c^2)/4$$)

0 = -$$\sqrt{3}$$c + ($$\sqrt{3}$$c2)/((2c)/2)

0 = -$$\sqrt{3}$$c + ($$\sqrt{3}$$c2)/c

0 = -$$\sqrt{3}$$c + ($$\sqrt{3}$$c

Did i plug my a into the wrong equation

16. May 13, 2009

Dick

It looks to me like you are plugging a=sqrt(3)*c back into the same equation you used to derive that relation. Plug it into dA/da, not dA/dc.

17. May 13, 2009

joemama69

ok so ..

Aa = P/2 - a - c - a2/(8$$\sqrt{c^2 - (a^2)/4}$$) + ($$\sqrt{c^2 - (a^2)/4}$$)/2 = 0, plug in a = $$\sqrt{3}$$c

0 = P/2 - $$\sqrt{3}$$c - c - 3c2/(8$$\sqrt{c^2 - (3c^2)/4}$$) + ($$\sqrt{c^2 - (3c^2)/4}$$)/2

0 = P/2 - $$\sqrt{3}$$c - c - 3c2/(4c) + c/4

0 = P/2 - $$\sqrt{3}$$c - c - 3c/4 + c/4

P = c(2$$\sqrt{3}$$ + 7/2)

c = P/(2$$\sqrt{3}$$ + 7/2)

18. May 13, 2009

Dick

Fine, except you made an arithmetic error. In problems like these you can have several choices of variables to eliminate, and several choices of which equation and variable to solve for first. Try and look ahead and figure out which choice will make your life easier.

19. May 13, 2009

joemama69

Where is my error @, i cant find it

20. May 13, 2009

Dick

Between the third line from the end and the next one.