Minimal surface area for a fixed volume

[lee123456789]

Homework Statement

a hut has to side walls a roof and back wall. its front is open. its total volume is 120m^3 fdetermine the miniumal surface area necessary for a sheet to be put over it

Homework Equations

The Attempt at a Solution

Attempt 2
V=xyz=120 z=120/xy
s = 2yz + xz + xy
s = 2y(120/xy) + x(120/xy) + xy s = (240/x) + (120/y) + xy
s' = y - 240/x^2 = 0
S' = x - 120/y^2 = 0

(X^2)y = 240
(y^2)x = 120
(X^2)y/(y^2)x = 240/120 = 2
2= x/y
x=2y
x=2y
2y * (y^2) = 2y^3
2y^3 = 120
y = 3.91
x = 7.82

Z = 120/xy
Z = 120/(3.91)(7.82) = 3.92
Z = 3.92

s'' = 480/x^3 = 480/7.82^3 = 1
s'' = 240/y^3 = 240/3.91^3 = 4
(X^2)y/(y^2)x = 240/120 = 2

(2^2) - (1)(4) = 0
(X^2)y/(y^2)x = 240/120 = 2 postive so is min
s = 2yz + xz + xy
2(3.91)(3.92) + (7.82)(3.92) + (3.91)(7.82) = 91.88m^3 surface area min

 

[RPinPA]

s = 2yz + xz + xy s = 2y(120/xy) + x(120/xy) + xy s = (240/x) + (120/y) + xy

A comment on formatting. This was a little hard to read. I understand the urge to compress things into a single line when you have so many equations to enter, but I prefer to put an implication symbol $\implies$ or $\iff$ or $\rightarrow$ between equations when doing that, to make it more clear where one equation stops and the next begins.

Even without the LaTeX editor, you could use the convention -> for this purpose. But people here prefer you use LaTeX. See link to "LaTeX Guide" at lower left corner.

Thus: $s = 2yz + xz + xy \implies s = 2y(120/xy) + x(120/xy) + xy \implies s = (240/x) + (120/y) + xy$

s' = y - 240/x^2 = 0 S' = x - 120/y^2 = 0

I guess you mean these to be the partials with respect to $x$ and $y$ respectively. Again, not clear.

x/y = 240/120 = 2
X=y

Oops. If x/y = 2 then x and y are not equal. You handled this equation correctly in "Attempt 2", which I will now skip to.

x=2y 2yx(y^2) = y^4

Where did that second equation come from? Which equation are you substituting into?

y^4 = 120
y = 3.30
x = 6.60

I get 3.31 as the 4-th root of 120. But I think there's something suspect in your algebra, as $x^2y \neq 240$ and $y^2x \neq 120$. These are not the solutions to your system of equations.

Your approach seems mostly correct, I just think there are careless algebra errors in both attempts.

 

[lee123456789]

A comment on formatting. This was a little hard to read. I understand the urge to compress things into a single line when you have so many equations to enter, but I prefer to put an implication symbol $\implies$ or $\iff$ or $\rightarrow$ between equations when doing that, to make it more clear where one equation stops and the next begins.

Even without the LaTeX editor, you could use the convention -> for this purpose. But people here prefer you use LaTeX. See link to "LaTeX Guide" at lower left corner.

Thus: $s = 2yz + xz + xy \implies s = 2y(120/xy) + x(120/xy) + xy \implies s = (240/x) + (120/y) + xy$I guess you mean these to be the partials with respect to $x$ and $y$ respectively. Again, not clear.
Oops. If x/y = 2 then x and y are not equal. You handled this equation correctly in "Attempt 2", which I will now skip to.Where did that second equation come from? Which equation are you substituting into?
I get 3.31 as the 4-th root of 120. But I think there's something suspect in your algebra, as $x^2y \neq 240$ and $y^2x \neq 120$. These are not the solutions to your system of equations.

Your approach seems mostly correct, I just think there are careless algebra errors in both attempts.

it was made up of two overal equations
i got it from these
s' = y - 240/x^2 = 0
S' = x - 120/y^2 = 0

(X^2)y = 240
(y^2)x = 120

then if 0.5X = y
2y subbed into equation = y^4
thought it would be easier than using 1/2x in the x equation

what but do i have to look atr if u don't mind me asking

 

[RPinPA]

2y subbed into equation = y^4

Please be more explicit in what you're doing.
$x^2 y = 240$.
Replace $x$ by $2y$
$4y^2 * y = 240$
I'm not seeing where that becomes $y^4 = 120$.

Please show your steps.

 

[lee123456789]

Please be more explicit in what you're doing.
$x^2 y = 240$.
Replace $x$ by $2y$
$4y^2 * y = 240$
I'm not seeing where that becomes $y^4 = 120$.

Please show your steps.

x=2y
$y^2 x= 120$.

$y^2 * 2y = 120$.
$2y^3 = 120$
$y^3 = 60$
root(60)
y = 3.91

2y = x
3.91 x 2 =x
x = 7.82i got muddle with my powers
can i sub both 2y into the $X^2 y = 240$ or the other way round x into the $y^2 x = 120$

Z = 120/xy
Z = 120/(3.91)(7.82) = 3.92
Z = 3.92

s'' = 480/x^3 = 480/7.82^3 = 1
s'' = 240/y^3 = 240/3.91^3 = 4
(X^2)y/(y^2)x = 240/120 = 2

(2^2) - (1)(4) = 0
(X^2)y/(y^2)x = 240/120 = 2 postive so is min
s = 2yz + xz + xy
2(3.91)(3.92) + (7.82)(3.92) + (3.91)(7.82) = 91.88m^3 surface area min

 

[haruspex]

for a sheet to be put over it

I'm not sure what this means. It sounds like we are putting a sheet over the whole shed, which would make the fact that the front is open is irrelevant. But I suspect that you are right, that it intends just the surface area of roof and three walls.

91.88m^3

You could get a bit more accuracy (your answer is not as accurate as you are effectively claiming) by keeping everything in an exact algebraic/rational form until the final step. I.e. from 2y³ = 120, just obtain that y=60^1/3 and leave it at that until you have combined the three dimensions back into the expression for total area and simplified.