Is something wrong with LaTeX?
Can you guys see it ok?
I found this one pretty hard..
A cuboid has volume of . The base of the cuboid is square with sides of length . The Surface area of the cuboid is
1) Show that
2) Find
3) Find the value of x which gives the smallest Surface Area of the cuboid, justifying your answer.
Please help! I have a real exam on this soon and the rest of Core 1 is easy except these stupid little questions..
Thanks guys!
[QUOTE=anthmoo;32716]I found this one pretty hard..
A cuboid has volume of . The base of the cuboid is square with sides of length . The Surface area of the cuboid is
1) Show that [quote]
Apparently, LaTex is down. Has been for a few days now.
Anyway. The volume of your cuboid could be represented as:
(x^2)y=8
y=8/x^2
A=2x^2+4x(8/x^2)=2x^2+(32/x)
Now, do that differentiate thing to find parts 2 and 3
[QUOTE=galactus;32718][QUOTE=anthmoo;32716]I found this one pretty hard..
A cuboid has volume of . The base of the cuboid is square with sides of length . The Surface area of the cuboid is
1) Show thatThanks!
Apparently, LaTex is down. Has been for a few days now.
Anyway. The volume of your cuboid could be represented as:
(x^2)y=8
y=8/x^2
A=2x^2+4x(8/x^2)=2x^2+(32/x)
Now, do that differentiate thing to find parts 2 and 3
I've differentiated it to 4x-32x^-2
but how do i find the smallest value for x to give the smalles surface area?
LaTeX is not working so I will translate this into ASCII
RonLI found this one pretty hard..
A cuboid has volume of 8 m^3. The base of the cuboid is square with sides of length x m . The Surface area of the cuboid is A m^2
1) Show that A=2x^2+32/x[/tex]
2) Find dA/dx
3) Find the value of x which gives the smallest Surface Area of the cuboid, justifying your answer.
Please help! I have a real exam on this soon and the rest of Core 1 is easy except these stupid little questions..
Thanks guys!
[quote=anthmoo;32719][quote=galactus;32718]The minimum surface area corresponds to the x which is a root of:
dA/dx=0,
so if your work is right you need to find the roots of:
4x-32x^-2=0,
and if there are more than one identify which corresponds to a minimum.
A root of dA/dx=0 corresponds to a minimum of A(x) if d^2A/dx^2<0 at the root.
RonL
Hello, anthmoo!
Let me walk through this one . . .
A cuboid has volume of 8 m³.
The base of the cuboid is square with sides of length x m.
The surface area of the cuboid is A m².
(1) Show that: .A .= .2x² + (32/x)
(2) Find: .dA/dx
(3) Find the value of x which gives the smallest Surface Area of the cuboid.Code:x * - - - * x/ / | * - - - * | | | |y | | | y| y| | | | * | | /x * - - - * x
We are given the volume: .V .= .x²y .= .8 . → . y .= .8/x² . [1]
The top and bottom have an area of: 2x²
The four sides have an area of: 4xy
The total surface area is: .A .= .2x² + 4xy . [2]
Substitute [1] into [2]: . A .= .2x² + 4x(8/x²)
Hence, we have: . A .= .2x² + (32/x) . (1)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We have: .A .= .2x² + 32x^{-1}
Then: .dA/dx .= .4x - 32x^{-2} . (2)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We have: . 4x - 32x^{-2} .= .0
Multiply by x²: . 4x³ - 32 .= .0
And we have: . 4x³ = 32 . → . x³ = 8 . → . x = 2 .(3)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Substitute into [1]: . y .= .8/2² . → . y = 2
As you may have suspected,
. . the cuboid with minimum surface area is a cube.