the length of a cuboid is twice its width...

bumblebee123

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can anyone help to explain this?

question: the length of a cuboid ( 2x ) is twice its width ( x )

the volume of the cuboid is 72 cm^3
the surface area of the cuboid is A cm^2

show that A = 4x^2 + ( 216/ x )

any help would be highly appreciated! :)
 
You don't know what the height is; call it y. (Or whatever you like!)

Write an equation that says the volume is 72 cm^3.

Write an equation that says the surface area is A cm^2.

Show them to us, and try substituting in the second equation from the first to eliminate y.

Then we can help.
 
volume = y x 2x x x = 2x^2 x y

72 = 2x^2 x y

surface area = 2( 2xy ) + 2( xy ) + 2 (2x^2 )

surface area= 4xy + 2xy + 4x^2

A = 4xy + 2xy + 4x^2
 
volume = y x 2x x x = 2x^2 x y

72 = 2x^2 x y

surface area = 2( 2xy ) + 2( xy ) + 2 (2x^2 )

surface area= 4xy + 2xy + 4x^2

A = 4xy + 2xy + 4x^2
You have like terms so combine them. You still have to get rid of the y. Also do not use x to mean multiplication.
 
Please, don't use "x" for both a variable and a multiplication sign! That is utterly unreadable. Use "*" as we traditionally do for multiplication in typing.

So you have 2x^2 y = 72, and S = 6xy + 4x^2 (which you got correctly, but didn't combine fully).

Now, as I suggested, solve the first for y and substitute that in the second.
 
Please, don't use "x" for both a variable and a multiplication sign! That is utterly unreadable. Use "*" as we traditionally do for multiplication in typing.

So you have 2x^2 y = 72, and S = 6xy + 4x^2 (which you got correctly, but didn't combine fully).

Now, as I suggested, solve the first for y and substitute that in the second.

I didn't know what else to use, sorry. also, how do I solve for Y if I don't know the value of X?
 
Do the same sort of thing you would do to solve something like 2x + 3y = 6 for y, which I presume you have done. In that case, you treat x as if it were just a number, and isolate y on one side (some people call this "making y the subject"). You don't need to know the value of x to find an expression for y. In this example, you would subtract 2x from both sides, then divide by 3, yielding y = (6 - 2x)/3. In your problem, all you need is one division.

As for using x for multiplication, I thought someone had mentioned to you the link in our submission guidelines telling how to type math: "Formatting Math as Text". Maybe it was someone else. Anyway, now you know.
 
Do the same sort of thing you would do to solve something like 2x + 3y = 6 for y, which I presume you have done. In that case, you treat x as if it were just a number, and isolate y on one side (some people call this "making y the subject"). You don't need to know the value of x to find an expression for y. In this example, you would subtract 2x from both sides, then divide by 3, yielding y = (6 - 2x)/3. In your problem, all you need is one division.

As for using x for multiplication, I thought someone had mentioned to you the link in our submission guidelines telling how to type math: "Formatting Math as Text". Maybe it was someone else. Anyway, now you know.


so i have two equations:

2x^2 y = 72

s = 6xy + 4x^2

if I change the first equation so that y = 72 - 2x^2

s = 6x ( 72 - 2x^2 ) + 4x^2

is this what I'm supposed to do??

( also thanks for the website link! )
 
so i have two equations:

2x^2 y = 72

s = 6xy + 4x^2

if I change the first equation so that y = 72 - 2x^2

No. To isolate y you need to divide both sides by \(\displaystyle 2x^2\) not subtract. (y is being multiplied by \(\displaystyle 2x^2 \)in the original equation so you need to do the opposite of multiplication which is division.)

\(\displaystyle y=\frac{72}{2x^2} =\frac{36}{x^2}\)

Continue from here...


s = 6x ( 72 - 2x^2 ) + 4x^2

is this what I'm supposed to do??

( also thanks for the website link! )
see comments in red
 
2x^2 y = 72

s = 6xy + 4x^2

if I change the first equation so that y = 72 - 2x^2
You should always think about what you are doing to both sides of an equation and actually do it. Here, you subtracted 2x^2 from the right side; but if you did that to both sides you would get

2x^2 y - 2x^2 = 72 - 2x^2

and the left side does not simplify to y!

Rather than subtract, you should have divided (as I told you!), because the goal is to undo a multiplication.
 
You should always think about what you are doing to both sides of an equation and actually do it. Here, you subtracted 2x^2 from the right side; but if you did that to both sides you would get

2x^2 y - 2x^2 = 72 - 2x^2

and the left side does not simplify to y!

Rather than subtract, you should have divided (as I told you!), because the goal is to undo a multiplication.

sorry, my bad, I don't know why I subtracted it
 
y =72 / 2x^2 = 36 / x^2

A = 6xy + 4x^2

A = 6x ( 36 / x^2 ) + 4x^2

A = ( 216x / 6x^3 ) + 4x^2
 
When you multiply a fraction (by 6x, here), you multiply only the numerator, since you are multiplying by the fraction 6x/1.

Multiplying both numerator and denominator just produces an equivalent fraction by multiplying by 1 (in the form (6x)/(6x)).

Try again.

As for the mistake you made, it is very common. You can avoid it by always thinking about what operation you are undoing; and you can catch the error by always actually doing what you say you are doing, rather than writing what you expect the next step to look like.
 
When you multiply a fraction (by 6x, here), you multiply only the numerator, since you are multiplying by the fraction 6x/1.

Multiplying both numerator and denominator just produces an equivalent fraction by multiplying by 1 (in the form (6x)/(6x)).

Try again.

As for the mistake you made, it is very common. You can avoid it by always thinking about what operation you are undoing; and you can catch the error by always actually doing what you say you are doing, rather than writing what you expect the next step to look like.

ah, makes sense

A = ( ( 6x/ 1) * ( 36/ x^2 ) ) + 4x^2
A = ( 216x / x^2 ) + 4x^2
A = ( 216 / x ) + 4x^2

thanks!
 
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