Can anyone explain questions 2 and 3 on problem set 2? Problem Set - http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/problem-set-2/MIT18_02SC_pset2.pdf Problem Set Solutions - http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/problem-set-2/MIT18_02SC_pset2sol.pdf

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Can anyone explain questions 2 and 3 on problem set 2? Problem Set - http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/problem-set-2/MIT18_02SC_pset2.pdf Problem Set Solutions - http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/problem-set-2/MIT18_02SC_pset2sol.pdf

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For problem 2, Lets assume we have produced p1 units of P1, p2 units of P2 and p3 units of p3. Given the ratios, to produce p1 units of P1, we need (1/6)p1 units of M1, (1/3)p1 units of M2 and (1/2)p1 units of M3, and so on for P1 and P2. Now, if we are to look at the total amount of units of M1 used, we find that it is (1/6)p1 units + (1/9)p2 units + (3/16)p3 units. This quantity is equal to 137. So we can write the euqation (1/6)p1 + (1/9)p2 + (3/16)p3 = 137. Similarly, if we track the amount of M2 and M3 used, we get the following equations- (1/3)p1 + (1/3)p2 +(5/16)p3 = 279 (1/2)p1 + (5/9)p2 + (1/2)p3 = 448 We can represent these three equations in the matrix form as \[\left[\begin{matrix}1/6 & 1/9&3/16 \\ 1/3 & 1/3&5/16\\1/2&5/9&1/2\end{matrix}\right]\left(\begin{matrix}p1 \\ p2\\p3\end{matrix}\right)=\left(\begin{matrix}137 \\ 279\\448\end{matrix}\right)\] Then you solve from there, by finding the inverse of the matrix etc as stated in the solution
That was my reasoning of the problem. I hope it makes sense :)
  • phi
I am not sure this will help... it depends on exactly what your question is.
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