The problem involves finding the length of a diagonal of a cuboid given its volume, surface area, and one edge length. The cuboid has a volume of 40 cm³, a total surface area of 100 cm², and one edge measuring 2 cm. The original poster expresses difficulty in approaching the problem logically during an exam setting.
Participants are exploring different interpretations of the problem, discussing the relationships between the dimensions, volume, and surface area. Some guidance has been offered regarding the use of different variable names for length, width, and height, and the need to set up equations based on the provided values.
There is a noted constraint regarding the interpretation of the dimensions, as the original poster has only one known edge length and is trying to derive the others from the volume and surface area equations. The discussion highlights the challenge of having multiple unknowns with limited information.
Oops. What is ##\frac {120} H * H##Physiona said:And then it'll equal:
4H + 120/H * H = 100.
The H will cancel, leaving:
4H + 120 = 100
Oh, is it 40H/H?tnich said:Oops. What is ##\frac {120} H * H##
Sorry, I meant what is ##\frac {40} H * H##
Sorry, I messed that up. Let's go back to post #27 where you say correctly thatPhysiona said:Is it 120H/H?
What is ##\frac {40} H * H##?Physiona said:4H + 80/H + 40/H * H = 100
OK, what is ##\frac H H##? It's not a trick question.Physiona said:Yes would it equal 40H/H?
Right. So now what does your equation 4H + 80/H + 40/H * H = 100 become?Physiona said:That's 1 isn't it?
Sorry, having a rough day.
So it's 40?
OK. Now solve it for H.Physiona said:4H + 80/H + 40 = 100
Try multiplying the whole equation by H.Physiona said:The fraction bit of 4H and 80/H is confusing. Do I multiply 4H by H and divide 80?
Nope. You need to multiply each term in the equation (on both sides) by H. Also, what is ##\frac H H##?Physiona said:Will I get 4H2 + 120 = 100
Right. Can you explain to me why if ##x + y + z = c##, then ##ax + ay + az = ac##?Physiona said:4H * H = 4H2
80/H = 80H/H or 80 I presume
40 * H = 40H
100 * H = 100H
Is this right?
Right. And why does that work?Physiona said:You're multiplying by a coefficient of a for all the terms aren't you?
If you start with ##x=3## and then say if you multiply both sides by a, then why is ##ax = 3a##?Physiona said:Because they all have a common like term in them? I'm not sure..
Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say ##x=3##. If you take that number and multiply it by a, then you get the same result whether you call it ##x## or 3. So you can say ##ax = 3a##. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.Physiona said:To make it balanced with the same coefficient. Right? As you multiply both sides by a..
Physiona said:To make it balanced with the same coefficient. Right? As you multiply both sides by a..
I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)tnich said:Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say ##x=3##. If you take that number and multiply it by a, then you get the same result whether you call it ##x## or 3. So you can say ##ax = 3a##. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.
So now, take your equation 4H + 80/H + 40 = 100 and multiply by H.
You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.Physiona said:I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)
You should get a single equation as a result.tnich said:You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.
Great! Now solve that for H.Physiona said:I get 4H2 + 80 + 40H = 100H
Go ahead.Physiona said:Okay thank you!
So do I collect like terms from both side:
4H2 + 80 = 100H - 40H
4H2+80= 60H
Hang on, can I solve it as a quadratic?
In the form of:
4H2-60H+80=0
I think so. Now that you have H, how are you going to find L?Physiona said:I end up with two solutions of:
13.520797289396
1.4792027106039
Is this correct?