Question #de470

1 Answer
Feb 21, 2017

Kilos of rambutan: #8#
Kilos of duku: #6#
Kilos of langsat: #4#

(see below for solution method)

Explanation:

Part a
Let #R# be the number of kilos of rambutan,
#K# be the number of kilos of duku (I used #D# for Determinant, so I didn't want to use it for "duku" as well), and
#L# be the number of kilos of langsat.

We are told the total cost (and individual per kilo costs); so:
#color(white)("XXX")0.75R+0.90K+0.60L=13.80#
Also the total weight:
#color(white)("XXX")R+K+L=18#
And that the difference of the total cost of the duku and langsat minus the cost of the rambutan:
#color(white)("XXX")-0.75R+0.90K+0.60L=1.80#

Part b
Setting this up as an augmented matrix:

#color(white)("XXX")Rcolor(white)("XXX")Kcolor(white)("XXX")Lcolor(white)("XXX")"constants"#

#( (0.75,0.90,0.60," | ",13.80), (1,1,1," | ",18), (-0.75,0.90,0.60," | ",1.80) )#

and using the (hopefully) standard designations for the derived matrices:
#M, M_R, M_K, M_L#

Crammer's Rule tells us that (for example):
#color(white)("XXX")R=("Determinant"(M_R))/("Determinant"(M))#

Here it is as evaluated in a spreadsheet:
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