Question 150550
Let {{{s}}} = lbs of soybean meal to be used
Let {{{c}}} = lbs of cornmeal to be used
Protein in the soybean meal = {{{.14s}}}
Protein in the cornmeal = {{{.07c}}}
In words:
(total lbs of protein in mixture) / (lbs of mixture) = % protein
{{{(.14s + .07c) / 280 = .1}}}
multiply both sides by {{{280}}}
(1) {{{.14s + .07c = 28}}}
Also known is {{{s + c = 280}}}
Multiply both sides by {{{.14}}}
(2) {{{.14s + .14c = 39.2}}}
Subtract (1) from (2)
{{{.07c = 11.2}}}
{{{c = 160}}} lbs
And since {{{s + c = 280}}}
{{{s + 160 = 280}}}
{{{s = 120}}} lbs
160 lbs of cornmeal are needed and 120 lbs of soybean meal are needed
check:
{{{(.14s + .07c) / 280 = .1}}}
{{{(.14*120 + .07*160) / 280 = .1}}}
{{{(16.8 + 11.2) / 280 = .1}}}
{{{28 / 280 = .1}}}
{{{28 = 28}}}
OK