According to the first scheme, three sets of equations
360/x+360/y=75
480/x+240/y=70
X= 12 y=8 can be obtained and substituted into the second scheme.
Scheme 2 600/x+ 120/y=65.
The second question, first look at the size box required for each scheme, which can be obtained by calculation.
Need a lot of boxes, need a few boxes.
Alternative 1: 30 45
Option 2: 50 15
Option 3: 40 30
Then set the packaging cost of small box as m, and the packaging cost of large box as m*( 1+k%).
Then the costs of the three schemes are as follows:
Scheme 1: 45 * m+30 * m * (1+k%) = 75 * m+30 * m * k%.
Scheme 2:15 * m+50 * m * (1+k%) = 65 * m+50 * m * k%.
Scheme 3: 30 * m+40 * m * (1+k%) = 70 * m+40 * m * k%.
Compare these three figures to see which scheme has the lowest cost.
Let the number of scheme 3 minus the number of scheme 2 =5*m- 10*m*k%.
Let the number of scheme 1 minus the number of scheme 2 = 10*m-20*m*k%.
When k is less than or equal to 25, both numbers are positive, so the cost of the second scheme is the lowest.
When k is greater than or equal to 55, both numbers are negative, so the cost of the second scheme is the highest.
In the comparison between Scheme I and Scheme III,
Scheme 1 minus Scheme 3 = 15*m- 10*m*k%.
When k is greater than or equal to 55, this number is positive, so the cost of the third scheme is the lowest.
To sum up, when k is less than or equal to 25, the cost of the second scheme is the lowest.
When k is greater than or equal to 55, the cost of scheme 3 is the lowest.