Question 185217
Let {{{s}}}= number of soccer balls to be made
Let {{{v}}}= number of volleyballs to be made
Let {{{P}}}= profit from both
given:
(1) {{{P = 5s + 4v}}}
(2) {{{(2/75)*s + (3/60)*v <= 500}}}
(3) {{{(3/75)*s + (2/60)*v <= 450}}}
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Multiply both sides of (2) and (3) by {{{300}}}
(2) {{{8s + 15v <= 150000}}}
(3) {{{12s + 10v <= 135000}}}
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I'll plot {{{s}}}(x-axis) and {{{v}}}(y-axis) (1),(2),and (3)
{{{ graph( 600, 600, -5, 20, -5, 20, (150 - 8x)/15, (135 - 12x)/10) }}} 
The axes are scaled to thousands
To find the intersection I'll multiply (1) by {{{3}}}
and (2) by {{{2}}} Then subtract (3) from (2)
(2) {{{24s + 45v = 450000}}}
(3) {{{-24s - 20v = -270000}}}
(4) {{{25v = 180000}}}
(4) {{{v = 7200}}}
Putting this back in (2)
Multiply both sides of (2) and (3) by {{{300}}}
(2) {{{8s + 15*7200 = 150000}}}
(4) {{{8s + 108000 = 150000}}}
(4) {{{8s = 42000}}}
(4) {{{s = 5250}}}
7200 volleyballs and 5250 soccer balls are made to maximize profits
The maximum profit is:
(1) {{{P[max] = 5s + 4v}}}
(1) {{{P[max] = 5*5250 + 4*7200}}}
(1) {{{P[max] = 26250 + 28800}}}
(1) {{{P[max] = 55050}}}
The maximum profit is $55,050
 



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