# Can someone check my maths please.Watch

#1
a rectangular box has two sides a back and top with volume of 120m^3 find minimum surface area of cloth that can go over the top.
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
Last edited by mlm1234; 3 weeks ago
0
3 weeks ago
#2
Seems pretty good apart from

2 = x/y
x=y

Haven't worked out the full solution though.
0
3 weeks ago
#3
(Original post by mlm1234)
a rectangular box has two sides a back and top with volume of 120m^3 find minimum surface area of cloth that can go over the top.
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 = x/y x=y

X^3 = 240 x = 6.21

s'' = 480/x^3 = 480/6.21^3 = 2
s'' = 240/y^3 = 240/6.21^3 = 1
(X^2)y/(y^2)x = 240/120 = 2

(1^2) - (2)(1) = -1
(X^2)y/(y^2)x = 240/120 = 2 postive so is min
s = 2yz + xz + xy 2(6.21)(3.11) + (6.21)(3.11) + (6.21^2) = 96.49m^3 surface area min
What do you mean by "over the top"? If you mean the total surface area then your answer is not correct. I haven't checked your working.
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#4
(Original post by BuryMathsTutor)
What do you mean by "over the top"? If you mean the total surface area then your answer is not correct. I haven't checked your working.
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
0
3 weeks ago
#5
(Original post by mlm1234)
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
You are correct up to x/y=2.

It would be helpful if you use LaTeX or put your equations on new lines or at least use a little more space between equations.
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