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What I have so far: Beyond the Galaxy We create trippy clothes. Such as (product #1:) shirts (x) and (product #2:) skirts (y). 30x+20y<= $3x+$5y<=$340
x is the number of items of product 1 so when you make the requirement that you must "Produce at least 30 of product #1", this means x >= 30 So you can produce x = 30, x = 31, x = 32, ... items of product 1 (but nothing less than 30)
So what do you get when you translate "Produce at least 20 of product #2"
you got it
Therefore, your system should be x >= 30 y >= 20 3x+5y <= 340
dude, thank you so much! :D
yw, the only thing left to do is graph do you know how to do that?
not with inuqualitites.. help?
to graph x >= 30, follow these steps Step 1) graph the vertical line x = 30 Step 2) make it a solid line Step 3) shade to the right of this solid line
To graph y >= 20, you * graph the horizontal line y = 20 * make it a solid line * shade above the solid line
to graph 3x+5y <= 340, you * graph the line 3x+5y = 340 * make it a solid line * shade below the line
all 3 regions overlap to form the final shaded region
how would I do 3x+5y <= 340?
plug in x = 0 to get 3x+5y = 340 3*0+5y = 340 5y = 340 y = 340/5 y = 68 So one point on the line 3x+5y=340 is (0,68)
Youre amazing! xD
thx, now you just need one more point to graph 3x+5y = 340
hold on, it didn't post part of the directions..
If Product #1 earns a profit of $10 per unit and Product #2 earns a profit of $15 per unit, find the combination of Product #1 and Product #2 that will maximize profit. A commercial that could be used to advertise your two products to the general public. You may include a slogan or jingle.
same problem or different one entirely?
I think.. It was just the directions continued
Let P = combined profit of products #1 and #2 So "If Product #1 earns a profit of $10 per unit and Product #2 earns a profit of $15 per unit", then P = 10x + 15y
The ideal situation is that there is an ordered pair (x,y) that is within the shaded region and makes P as large as possible
It turns out that this ordered pair is a vertex of the shaded region ie, it is formed by the intersection of two of the lines
here is the xy axis |dw:1358723166719:dw|
now here I'm going to add on the graphs of x = 30, y = 20, and |dw:1358723212768:dw|
the shaded region is here |dw:1358723376880:dw| and it includes the surrounding (triangular) boundary
x would be (30,0) and y would be (0,20) right?
what do you mean
for x>=30 I would graph (0,30) ?
i mean (30,0)
yes you would use any points with an x coordinate of 30
were you able to find the points of intersection?
still working on it
I have the points so far
So I connect them and shade the middle right?
those points you got look like they need more points to be able to draw the 3 lines
take a look at what I posted and see if that makes sense or not
It wont open
odd, it should work with geogebra your file worked
So I need to (30,0) and (30,4)
yes 2 points per line
the screen shot worked (:
ok great, the upper horizontal line should be diagonal like in the pic i posted
so how do I fix that?
make it go through the points (0,68) and (30,50)
Ive never used GeoGebra before
or you can simply type in 3x+5y = 340 into the input bar and it will graph the line 3x+5y = 340 for you
ok you got all 3 lines nailed down correctly
the shaded region will be the triangle in the middle and it will include the boundary
Ok awesome! ..how do I shade it?
you can use geogebra to do the shading, or you can use paint like I just did (I used the paint bucket) to fill in the correct region
Thank you so much!
sure thing, you can then easily use geogebra to find the points of intersection then you need to see which points give you the max profit
So the shaded region would be P = 10x + 15y, right?
no that has nothing to do with the shaded region once you find the vertices, you use P = 10x + 15y to test each one
no it won't, the shaded region is done
what are the points of intersection
Oh! I plug in the x and y coordinates into P = 10x + 15y
then you plug them into P = 10x + 15y
Okays thank you (:
the ordered pair that gives you the largest value of P is the winner