ATTENTION I ALREADY KNOW THE ANSWER I JUST WANT SOMEONE TO HELP ME WORK THE PROBLEM STEP BY STEP An artist is creating a mosaic that cannot be larger than the space allotted which is 4 feet tall and 6 feet wide. The mosaic must be at least 3 feet tall and 5 feet wide. The tiles in the mosaic have words written on them and the artist wants the words to all be horizontal in the final mosaic. The word tiles come in two sizes: The smaller tiles are 4 inches tall and 4 inches wide, while the large tiles are 6 inches tall and 12 inches wide. If the small tiles cost $3.50 each and the larger tiles cost $4.50 each, how many of each should be used to minimize the cost? What is the minimum cost
@phi can you please help
we need some "relations" that tell us how many big tiles we can use and how many small tiles we can use.
for example, the big tiles (measured in feet) or 0.5 tall and 1 wide if the mosaic must be 3 feet tall, how many big tiles do we need to make a height of 3 feet?
6?
yes.
Ok, what is next?
well this is what I got as the answer 0 small tile, 6 large tiles; minimum cost $27 so now i just need to know how to work the problem all together
This is a problem in linear programming. The area of a big block is ½ sq ft the area of a small block is 1/9 sq ft if we have x small blocks and y big blocks, the area is x/9 + y/2 this area must be larger than the 3 x 5 ft min= 15 sq ft but smaller than the large region 4 x 6 = 24 \[ 15≤ \frac{x}{9}+ \frac{y}{2} ≤ 24 \] we would plot those two lines as a start
ok, I am with you
we will get an "allowed region" between the two lines. we next need to plot lines showing the allowed max number of blocks.
ok, can you show me?please
Here are the two lines
I think the max is 6
nevermind i think i got it thank you
ok
Here is the plot with the intersections at points A, B , C and D we optimum point will be one of these. We evaluate the cost of each of these points
the cost of point A (0,48) (where 48 is the number of big blocks) is 4.50*48= $216 the cost of point B is 4.50*30= $135 the cost of point C (135,0) is 135*3.50= $472.50 and the cost of point D (216,0) is 216*3.50= $756
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