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Mathematics 17 Online
OpenStudy (anonymous):

Find max and min

OpenStudy (anonymous):

Can someone help me find P3_min and P3_max \[\left[\begin{matrix}100 & 100&100 \\ 200 & 100&0\\ 100&0&200\end{matrix}\right]*\left(\begin{matrix}d1 \\ d2\\d3\end{matrix}\right)=\left(\begin{matrix}100 \\ 100\\P3\end{matrix}\right)\] Where d1,d2,d3≥0

OpenStudy (anonymous):

And d1+d2+d3=1

OpenStudy (anonymous):

Anyone?

OpenStudy (anonymous):

@perl Can you help me?

OpenStudy (perl):

this is a max and min problem?

OpenStudy (anonymous):

Yes I need to find min and max P3

OpenStudy (perl):

why is it called min and max

OpenStudy (perl):

we can solve the augmented matrix

OpenStudy (anonymous):

I know the result is 0<P3<150. But I cant find at way to show it

OpenStudy (anonymous):

If we /100 we get \[\left[\begin{matrix}1 & 1&1 \\ 2 & 1&0\\ 1&0&2\end{matrix}\right]*\left(\begin{matrix}d1 \\ d2\\d3\end{matrix}\right)=\left(\begin{matrix}1 \\ 1\\P3/100\end{matrix}\right)\]

OpenStudy (perl):

whats the name of this class

OpenStudy (anonymous):

Math finance..

OpenStudy (perl):

maybe if you could write more details of the chapter and stuff i could google it.

OpenStudy (usukidoll):

perlll!!!!! lol

OpenStudy (perl):

we have the three equations d1 + d2 + d3 = 1 2d1 + d2 = 1 d1 + 2d3 = p3/100 d2 = 1 - 2*d1, d3 = 1/2* (p3/100 - d1) so plugging that into our first equation we have d1 + (1-2*d1) + 1/2 (p3/100 - d1) = 1

OpenStudy (usukidoll):

looks like an elementary linear algebra problem

OpenStudy (anonymous):

OpenStudy (anonymous):

@UsukiDoll I agree :) But I cant sole it.

OpenStudy (usukidoll):

umm reduce row echelon form it?

OpenStudy (usukidoll):

ah... ok first multiply the d column with that 3x3 matrix and then reduce row echelon it.

OpenStudy (usukidoll):

it's not in REF form...hmm need an upper triangular matrix

OpenStudy (usukidoll):

|dw:1381393016138:dw| you see the triangles? that's the main diagonal...anything below the main diagonal must be 0 for it to be in REF form.

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