+130555 Let x be the number of "X" cabinets and y be the number of "Y" cabinets.
And we are told the following.......
x / y ≥ 2/3 → 3x ≥ 2y → y ≤ ( 3/2 ) x
100x + 200y ≤ 1400 ..... this is the cost constraint
.6x + .8y ≤ 7.2 .......this is the constraint on the square meters
We also need two more contraints: x ≥ 0 and y ≥ 0, since we can't have a negative number of cabinets!!!
And we want to maximize the cubic meters of file storage .......we can just call this..... .8x + 1.2y
Have a look at the graph of the inequalities, here.........https://www.desmos.com/calculator/rxzwqo3dnw
The maximum for the objective function occurs at a corner point in the feasible region......the graph shows that there are two "whole number" corner points at (8, 3) and (12, 0)....another corner point occurs at (3.5, 5.25)....but.....we can't buy "partial" numbers of cabinets.....!!!
Notice, at (8, 3), the objective function = .8(8) + 1.2(3) = 10
At (12, 0), the objective function = .8(12) + 1.2(0) = 9.6
It looks like the best option for maximum storage under the given constraints is to purchase 8 of the "X" cabinets and 3 of the "Y" cabinets
Sorry....I don't know a second method.....
+130555 The equation is:
(y - 5)^2 / 9 - (x - 4)^2/ 4 = 1 multiply both sides by 36
4(y - 5)^2 - 9(x - 4)^2 = 36 expand
4(y^2 - 10y + 25) - 9(x^2 - 8x + 16) = 36 simplify
4y^2 - 40y + 100 - 9x^2 + 72x - 144 = 36
4y^2 - 40y -9x^2 + 72x - 80 = 0 to put in the proper form, multiply by -1 and rearrange
9x^2 - 4y^2 -72x + 40y + 80 = 0
Here's the graph of both forms......notice that they "overlay" each other......so......they are the same graph!!
https://www.desmos.com/calculator/i03ahreyrr
