How would I write this equation?

S

Saraug27

New member
Joined
Oct 22, 2012
Messages
3
Hi! I am an avid math student, and, after much work, decided I could not in any way with my current knowledge figure out how to equate this.
A converts to B at a ratio of 20a=1b, and only from A to B, not B to A.
I need 3 B and 5 A to make C.

Is there a way I could factor in both my current amount of B and my current amount of A to figure out how many conversions of A:B I should make to receive the maximum amount of C?

I do understand that I could do the math every time to figure out the correct numbers, but would there be an equation for it?
 
If I understand this correctly, you have \(\displaystyle B_0\) of "B" and \(\displaystyle A_0\) of "A" and can convert "A" into "B", at the rate of 1 unit of "B" for each 20 of "A". Let x be the amount of "A" that is converted to "B". That leaves \(\displaystyle A_0- x\) while giving \(\displaystyle B_0+ x/20\) of B. Since you "need 3 B and 5 A to make C", the amount of "C" made is \(\displaystyle B/3+ A/5= (B_0+ x/20)/3+ (A_0- x)/5= (B_0/3+ A_0/5)+ (1/60- 1/5)x= (B_0/3+ A_0/5)- (11/60)x\). That is linear and, because the coefficient of x is negative, that will be maximum when x= 0.
 
Last edited by a moderator:
[ֹtex] is a system command for LaTex, a program that renders nice math typesetting. The actual LaTex codes are typed next (each code begins with \). The closing tag [ֹ/tex] goes at the end.

\(\displaystyle \dfrac{x}{x + 4} = e^{x}\)


To see the actual coding, you may right-click on these renderings and then follow menus "Show math as Tex commands" (or click the [Reply to Post] button). :cool:


Ah, yes. Sometimes posters make a typographical error in the coding, and then strange renderings appear in the post. I'll fix that.
 
Last edited:
Saraug27 said:
Thank you so much! One more question, what does \(\displaystyle mean? :D\)
Click to expand...
\(\displaystyle
It has no meaning. It is a marker for the site to render the following text using LaTex, a formatting language. However, an end marker is required as well, and Halls forgot it.

Had he put the end marker on what you would have seen is:

\(\displaystyle B/3+ A/5= (B_0+ x/20)/3+ (A_0- x)/5= (B_0/3+ A_0/5)+ (1/60- 1/5)x= (B_0/3+ A_0/5)- (11/60)x.\)\)
 
Last edited:
You must log in or register to reply here.
Top