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Oliga [24]
3 years ago
10

I NEED HELP PLEASE, THANKS! :)

Mathematics
1 answer:
Alenkasestr [34]3 years ago
3 0

Answer:

Option A

Step-by-step explanation:

This is a great question!

The first thing we want to do here is to graph the system of " inequalities " -

\begin{bmatrix}2x+9y\le \:100\\ 9x+y\le \:54\end{bmatrix}

As x is ≥ 0, respectively y ≥ 0, it makes things a lot simpler, as it kind of restricts us to quadrant 1, but not entirely.

First change each inequality to " slope - intercept " form, as such -

2x + 10y \leq 100,\\10y \leq 100 - 2x,\\y \leq - 1 / 5x + 10\\----------\\9x + y \leq 54,\\y \leq - 9x + 54

The graphed solutions would be in the attachment. I have colored the region with which the two intersect, and the fact that we are limited to quadrant 1. In this colored region there are 3 major points, ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ). We have to determine which of these are are maximum points, as the minimum are the same. Substitute these values ( ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ) ) into the equation f( x, y ) = 10x + 4y,

f ( x, y ) = 10( 5 ) + 4( 9 ),\\f ( x, y ) = 86 -\\f ( x, y ) = 10( 0 ) + 4( 10 ),\\f ( x, y ) = 40 -\\f ( x, y ) = 10( 6 ) + 4( 0 ),\\f ( x, y ) = 60

( 5, 9 ) resulted in the greatest amount, 86. That would make it our maximum point -

<u><em>Solution = Option A</em></u>

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