Respuesta :
From the information that the question gives us, we have that models:
EZ weights 10 pounds, Box = 10 cubic-foot, represented by letter E
Mini weights 20 pounds, Box = 8 cubic-foot, represented by letter M
Hefty weights 60 pounds, Box = 28 cubic-foot, represented by letter H
The van can hold on a maximum of 440 pounds and has 296 cubic-feet. We have the following equations, to get the optimal solution:
For the weight:
[tex]10E+20M+60H=440[/tex]For the volume:
[tex]10E+8M+28H=296[/tex][tex]\begin{gathered} 10E+20M+60H=440 \\ -(10E+8M+28H)=296 \end{gathered}[/tex]We subtract the two equations to get:
[tex]12M+32H=144[/tex]Divide the above equation by 4, to simplify the equation:
[tex]3M+8H=36[/tex]Now we express M in terms of H:
[tex]\begin{gathered} 3M=36-8H \\ M=\frac{36-8H}{3} \end{gathered}[/tex]Now we start looking for integer solutions for the above equation, given that we want to maximize H, we can look for integer solutions starting with the largest possible, which is 4. But H=4 does not give an integer solution for M. We can try with H = 3
[tex]\begin{gathered} M=\frac{36-8\cdot3}{3} \\ M=4 \end{gathered}[/tex]Great, now that we have the number of M (Mini models), we can find the number of Hefty models and EZ models,
For H (Hefty models) we replace M in 3M+8H=36, and solve for H:
[tex]\begin{gathered} 3\cdot4+8H=36 \\ 12+8H=36 \\ 8H=36-12 \\ H=\frac{24}{8}=3 \end{gathered}[/tex]To find E, we can take the first equation 10E+20M+60H=440, to express E in terms of H and M, replace and solve:
[tex]\begin{gathered} 10E+20M+60H=440 \\ 10E=440-20M-60H \\ E=44-2M-6H \\ E=44-2\cdot4-6\cdot3=18 \end{gathered}[/tex]M = 4
H = 3
E = 18
