Question

The sum of the perimeters of an equilateral triangle and a square is 13. Find the dimensions of the triangle and the square that produce a minimum total area.

Answer 1

wow... ok

Answer 2

so du u know formula for area of square and area of triangle @akt_tippit ?

Answer 3

yes

Answer 4

sweet, so somehow we gotta make this a simultaneous eqn or a differential eqn...
so Area of square = s^2, yeah?
and Area of Triangle = 1/2 * base * ht, yeah?
and equilateral triangle soo... gimme a sec

Answer 5

Area of equilateral Triangle =
\[A = \frac {\sqrt 3} 4 \times base^2\]

Answer 6

agreed? @akt_tippit ?

Answer 7

agreed!

Answer 8

sweet, so now we hafta put in the perimeter side of this into the combined area equation, so we're only dealin with one equation that way... sooooooo

Answer 9

perimeter of an equilateral triangle = base + base + base = 3b
perimeter of a square = side + side + side + side = 4s

so if perimeter = 13 (total)
3b + 4s = 13

still with me man?

Answer 10

and combined area eqn is A square + Area triangle
Total Area = s^2 + sqrt 3/4 * b^2

Answer 11

so i think that's our 2 equations rite there...

Answer 12

so we solve for b and then put that equation into the total area equation

Answer 13

??

Answer 14

yep, sounds like a plan man, u go ahead and i'll watch an see how it goes?

Answer 15

jus lemme know if u want help with anything, k?

Answer 16

okay, so when you plug b in it becomes Square-root3/4(13-4s/3)^2+s^2 then square root 3/4(169/9 - 16s^2/9) + s^2? (I don't think I squared it right though) then 169square-root3/36 -16square-root 3 s^2/36 +s^2? I think I messed up somewhere

Answer 17

k heres what i got so far:

Answer 18

so \[3b + 4s = 13\] which means
\[3b = 13 - 4s\] so \[\large b = \frac {13 - 4s}3\]

\[\large Total~Area = s^2 + (\frac {\sqrt 3}4 \times b^2)\] so if
\[\large b = \frac {13 - 4s}3\] then
\[\large Total~Area = s^2 + (\frac {\sqrt 3}4 \times (\frac {13 - 4s}3)^2)\]

so the hard part: expand out the b^2
\[(\frac {13 - 4s}3)^2\]
\[(\frac {13 - 4s}3)(\frac {13 - 4s}3)\] so use FOIL and u should get
\[\frac {16s^2 - 104s +169}9\]
now put that back into the Total Area Eqn

\[\large Total~Area = s^2 + [\frac {\sqrt 3}4 \times (\frac {13 - 4s}3)^2]\]
\[\large Total~Area = s^2 + [\frac {\sqrt 3}4 \times (\frac {16s^2 - 104s +169}9)]\]
\[\large Total~Area = s^2 + [\frac {\sqrt 3}4 \times (\frac {16s^2 - 104s +169}9)]\]

Answer 19

\[\large Total~Area = s^2 + [\frac {\sqrt 3}4 \times (\frac {16s^2 - 104s +169}9)]\]
\[\large Total~Area = s^2 + [ (\frac {\sqrt 3 \times (16s^2 - 104s +169)}{4 \times 9})]\]
\[\large Total~Area = s^2 + [ (\frac {\sqrt 3 \times (16s^2 - 104s +169)}{4 \times 9})]\]
\[\large Total~Area = s^2 + [ \frac { 16s^2 - 104s +169}{\sqrt 3 \times 12}]\]

wow... maybe we should have solved for s in the first eqn ;D,

anyways, thats the equation hay
now iwe differentiate to find zero

Answer 20

so expanded form to put into y = ax^2 + bx + c
\[\large A = \frac {4s^2}{3 \sqrt3} + s^2- \frac{26 s}{3 \sqrt3} + \frac {169}{12 \sqrt3}\]
\[\large A = \frac {4+ 3 \sqrt3 }{3 \sqrt3}s^2 - \frac{26 s}{3 \sqrt3} + \frac {169}{12 \sqrt3}\]
so that can be solved by using quadratic equation:
\[\large y = ax^2 + bx + c\]
\[\large y = \frac {4+ 3 \sqrt3 }{3 \sqrt3}x^2 - \frac{26 x}{3 \sqrt3} + \frac {169}{12 \sqrt3}\]

us quadratic equation:
\[\huge x = \frac {-b± \sqrt{b^2-4 ac}}{2a}\]

now plug in your values for a, b and c above and you'll get
x = 1.4136 ish
so x is s
so a square with a side length of 1.4136

Answer 21

so use that value of s to work out b, so if u go back to
3b + 4s = 13
u can solve for b from here
those r ur dimensions for min area @akt_tippit

Answer 22

Wow Jack, that was a LOT of work

Answer 23

aww thanks wolf!
OP ran away... was kinda sad for a bit, love feelin appreciated so cheers man!

Answer 24

okay and cheers to you Jack
gee at least the OP could have left you a medal.

Answer 25

all good, my bad, maybe i should have rounded heaps earlier, thos sqrt signs are kinda scary ;D

Answer 26

LOL
well see ya around here no doubt

Answer 27

4 sure man, slaters!

Answer 28

How many hours did you spend on this thread?

Answer 29

Oh nevermind I see the timestamp. ~1 hour.

Answer 30

Good LaTeX skills right there. lol

Answer 31

cheers man!

Answer 32

jus hope OP comes back so is not 1 hr wasted xD

Answer 33

That was awesome! I was so frustrated with it and you helped out so much! thanks Jack!!!