Question

Please PLEASE PLEASE! Help me out! I would really appreciate if if someone helped me with a few questions about this...
4x + 3y ≥ 30
x + 3y ≥ 21
x ≥ 0, y ≥ 0

Minimum for
C = 5x + 8y

Answer 1

This is a classic linear optimization problem. The idea is that we're given a feasible region, described by a series of linear inequalities, and we need to find the point which maximizes our linear objective function, in this case \(C=5x+8y\). The neat property you should've learned is that maxima, points that maximize \(C\), lie at the vertices (or corners) of our feasible region i.e. where the linear inequalities form corners. Do you know what to do now?

Answer 2

Maybe step by step would be little better

Answer 3

I have more questions like this and I'd like to see the step by step of how to do them

Answer 4

I have to
Rewrite the constraints in slope-intercept form.
List all vertices of the feasible region as ordered pairs.
List all vertices of the feasible region as ordered pairs.
List the maximum or minimum amount, including the x, and y-value, of the objective function.

Answer 5

woops, I meant minimize and minima.

http://www.sophia.org/linear-programming/linear-programming--5-tutorial?pathway=systems-of-linear-equations-and-inequalities--2

Those inequalities *constrain* (enclose) our feasible region, so they're called *constraints*. By rewriting them in slope-intercept form, they merely want you to manipulate them until you have \(y\) all alone on the left-hand side with the right-hand side in the form \(mx+b\). So, for example, to rewrite our first constraint, \(4x+3y\ge30\), we can proceed as follows:$$4x+3y\ge30\\3y\ge-4x+30\\y\ge-\frac43x+10$$

The vertices of our feasible region are those corners I mentioned before. Since we're dealing with linear inequalities for our constraints, these corners are where the lines meet/intersect. Do you know how to solve for where two lines intersect?

Answer 6

I'm pretty sure

Answer 7

well, yes I do.

Answer 8

Well, let's convert our other constraint \(x+3y\ge21\) into slope-intercept form first:$$x+3y\ge21\\3y\ge-x+21\\y\ge-\frac13x+7$$Great, now we have both of our constraints solved for \(y\). To find where the lines intersect, we merely need to set \(y=y\) and substitute on either side:$$-\frac43x+10=-\frac13x+7\\10-7=-\frac13x+\frac43x\\3=x$$To determine the \(y\), we merely plug in:$$y=-\frac13(3)+7=-1+7=6$$So we've determined a vertex exists at (3,6).

Answer 9

ok

Answer 10

Thank you!