- #36
Physiona
- 131
- 9
That's 1 isn't it?
Sorry, having a rough day.
So it's 40?
Sorry, having a rough day.
So it's 40?
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?
Try them both and see what you get.Physiona said:Substitute it in the equation L.
(There's two numbers for H however..)
Notice anything odd/interesting about that?Physiona said:For H1= 13.520797289..
I got L1= 1.479202711
For H2= 1.4792027106039
I got L2= 13.52079729
Yes they are similar!tnich said:Notice anything odd/interesting about that?
There's a much easier way of solving this. Divide both sides of both equations by 2 to obtain:Physiona said:Yes I know, however I have unknowns,
So you'd make the surface area equation equal to the answer along with the Unknown coefficients.
2(2*H+L*2+L*H) = 100cm3
And the volume:
2*L*H =40cm3
Physiona said:Yes they are similar!
I presume I use one set of them into substituting it in the diagonal length formula.
Thank you so much, you have literally taught me this question in hours which my teacher can't even teach in 3 years!
I give you all my thanks, and thank you for guiding me till the very last.
Yes thank you for your guidance. As said before, this question came in my exam and I did struggle as I judged the question really quickly into thinking it's hard, however after multiple attempts and solutions I figured it. Thank you for the support though.Ray Vickson said:Not only are they similar, they just swap the variables. After all, YOU know what you mean by length and height, but the model does not! It just knows that one of them should be about 1.48 and the other should be about 13.52
Oh okay. May I ask how do you obtain the "Subtract 2 times Eqn. 2"? Do you multiply LH =20 by 2, and then subtract?Chestermiller said:There's a much easier way of solving this. Divide both sides of both equations by 2 to obtain:
$$2H+2L+LH=50\tag{1}$$
$$LH=20\tag{2}$$Subtract Eqn.2 from Eqn. 1 to yield: $$2(L+H)=30\tag{3}$$or$$L+H=15\tag{4}$$Square this equation to obtain:
$$(L+H)^2=L^2+2LH+H^2=225\tag{5}$$ Subtract 2 times Eqn. 2 from this equation to obtain:$$L^2+H^2=225-40=185\tag{6}$$Add ##W^2=4## to this equation to obtain:
$$L^2+W^2+H^2=185+4=189\tag{7}$$
Sure.Physiona said:Oh okay. May I ask how do you obtain the "Subtract 2 times Eqn. 2"? Do you multiply LH =20 by 2, and then subtract?