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fullonoob

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There is a steel rectangular prism, of dimensions 3x , x and h. The prism has volume of 4374m^3. Find the dimensions of the recyangular prism so that the minimum amount of steel is used.

Topic : Applications of Max and Min

Explain please.
No skipping steps :vcross:
 

shady145

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first draw a picture of a rectangular prism
next label the sides 3x, x, h
next, write out the formula for the volume of a rectangular prism... volume = b(times)l(times)h
so in this case V=3x(times)x(times)h=3hx^2
which was given before, V=4374, therefore 3hx^2=4374 call this equation one
now find an expression for the area of steel used
A=2(3xh + xh + 3x^2)
A=2(4xh + 3x^2) equation 2
from equation 1 find an expression for h
h=4374/x^2
sub that into eq 2,
then you will get a formula that has x's and no h's, call this equation, equation 3
differentiate this new formula after simplifying
solve it for when the first derivative is equal to zero
then differentiate it again to test for max or min
after you find the x value put it in the area formula (equation 3)

also you should post this in the MX1 section under NSW HSC

woops didnt read question properly =/
 
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cutemouse

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There is a steel rectangular prism, of dimensions 3x , x and h. The prism has volume of 4374m^3. Find the dimensions of the recyangular prism so that the minimum amount of steel is used.
3x.x.h=4374

h = 1458/x^2

Surface area, SA = 2*(3x.x + x.h + 3x.h) = 2*(3x^2 + 1458/x+ 4374/x) = 2(3x^2+5832/x)

SA' = 6x - 5832/x^2
SA'' = 6 + 11664/x^3

For a possible max/min SA'=0

6x= 5832/x^2

x^3=972
x=9.9 (to 1 dp)

When x=9.9 SA''=6>0 ==> Min SA when x=9.9

etc
 

fullonoob

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Well i did it the same way. It doesn't work out. You get x = 14.28...
That's the exact same thing i did. The real answers are 27 by 9 by 18 (dimensions)
 

The Nomad

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Oh dear, oh dear, oh dear. The rectangular prism is made of steel, so it doesn't require the minimum surface area, it requires the minimum sum of side lengths.

Let S = sum of side lengths.

S = 4(3x + x + h) = 4(4x + 1458/x^2)
S' = 4(4 - 2916/x^3)

For S' = 0, x^3 = 2916/4 = 729, x = 9. You can do the testing part. Enjoy.
 

cutemouse

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It's kind of a stupid question then because it doesn't define or give a diagram as to what we need to actually find/solve.
 

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