Question

Maximize Z = 2a + b – 3c + 5d
Subjected to;
a + 7b + 3c + 7d ≤ 46
3a – b + c +2d ≤ 8
2a + 3b – c + d ≤ 10
Where,
a, b, c, and d all are ≥ 0

Answer 1

Solve this problem by the method of Lagrange multipliers. We have

$\begin{cases} 2a+b-3c+5d\to\max\\ a+7b+3c+7d-46\le 0\\ 3a-b+c+2d-8\le 0\\ 2a+3b-c+d-10\le 0\\ -a\le 0\\ -b\le 0\\ -c\le 0\\ -d\le 0 \end{cases}$

Then $L(a,b,c,d,\lambda)=\lambda_0(2a+b-3c+5d)+$

$+\lambda_1(a+7b+3c+7d-46)+\lambda_2(3a-b+c+2d-8)+$

$+\lambda_3(2a+3b-c+d-10)-\lambda_4a-\lambda_5b-\lambda_6c-\lambda_7d$, where $\lambda=(\lambda_0,\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5,\lambda_6,\lambda_7)$

If $(a,b,c,d)$ is a point of local maximum, we have that there is $\lambda\neq 0$ such that

$\lambda_0\ge 0,$

$\lambda_1\le0, \lambda_2\le 0, \lambda_3\le 0, \lambda_4\le 0, \lambda_5\le 0, \lambda_6\le 0, \lambda_7\le 0$

$0=L'_a(a,b,c,d,\lambda)=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$

$0=L'_b(a,b,c,d,\lambda)=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$

$0=L'_c(a,b,c,d,\lambda)=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$

$0=L'_d(a,b,c,d,\lambda)=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$

$\lambda_1(a+7b+3c+7d-46)=0$

$\lambda_2(3a-b+c+2d-8)=0$

$\lambda_3(2a+3b-c+d-10)=0$

$\lambda_4a=0$

$\lambda_5b=0$

$\lambda_6c=0$

$\lambda_7d=0$

We consider the following cases:

1)$\lambda_0=0$

1.1)$\lambda_1=0$

1.1.1)$\lambda_2=0$

1.1.1.1)$\lambda_3=0$

Then from the equations

$0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$

$0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$

$0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$

$0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$

we obtain $\lambda=0$. This case does not satisfy us.

1.1.1.2)$\lambda_3<0$

From the equation

$0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$

we obtain that $\lambda_6>0$. This case does not satisfy us.

1.1.2)$\lambda_2<0$

From the equations

$0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$

$0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$

we obtain

$\lambda_4=3\lambda_2+2\lambda_3<0, \lambda_7=2\lambda_2+\lambda_3<0$

From the equations

$\lambda_4a=0, \lambda_7d=0, \lambda_2(3a-b+c+2d-8)=0$

we obtain

$a=0, d=0, 3a-b+c+2d-8=0$ , that is $a=d=0, c=b+8$

From the equations $\lambda_3(3b-c-10)=0, \lambda_5b=0, \lambda_6c=0$ we obtain $2\lambda_3(b-9)=0, \lambda_5b=0, \lambda_6(b+8)=0$

Since every system of two equations in $b-9=0, b=0, b+8=0$ has no solutions, we obtain that at least two numbers in $\{\lambda_3, \lambda_5, \lambda_6\}$ are zeros.

Then the equations

$0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$

$0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$

imply $\lambda_2=0$ . But it is contradicts with our case $\lambda_2<0$.

1.2)$\lambda_1<0$

From the equations

$0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$

$0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$

we obtain

$\lambda_4=\lambda_1+3\lambda_2+2\lambda_3<0$ and $\lambda_7=7\lambda_1+2\lambda_2+\lambda_3<0$

From the equations

$\lambda_1(a+7b+3c+7d-46)=0, \lambda_4a=0, \lambda_7d=0$

we obtain $a+7b+3c+7d-46=0, a=0, d=0$, that is $a=d=0, b=\frac{46-3c}{7}$

From the equations $\lambda_2(3a-b+c+2d-8)=0, \lambda_3(2a+3b-c+d-10)=0,$

$\lambda_5b=0, \lambda_6c=0$ we obtain

$\lambda_6c=0, \frac{3\lambda_5}{7}\left(\frac{46}{3}-c\right)=0,$

$\frac{10\lambda_2}{7}\left(c-\frac{51}{5}\right)=0, \frac{16\lambda_3}{7}\left(\frac{17}{4}-c\right)=0$

Since every system of two equations from $c=0, \frac{46}{3}-c=0, c-\frac{51}{5}=0, \frac{17}{4}-c=0$ has no solutions, we obtain that at least three numbers from $\{\lambda_6,\lambda_5,\lambda_2,\lambda_3\}$ are zeros. Then the equations

$0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$

$0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$

imply $\lambda_1=0$. It contradicts with our case $\lambda_1<0$.

2)$\lambda_0>0$. Let $\lambda_0=1$.

2.1)$\lambda_1=0$

2.1.1)$\lambda_2=0$

2.1.1.1)$\lambda_3=0$

Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $\lambda_4=2\lambda_0>0$. It does not satisfy us.

2.1.1.2)$\lambda_3<0$

From $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $0\ge\lambda_6=-3\lambda_0-\lambda_3=-3-\lambda_3$ and $0\ge\lambda_7=5\lambda_0+\lambda_3=5+\lambda_3$. System $-3-\lambda_3\le 0$, $5+\lambda_3\le 0$ does not have solutions.

2.1.2)$\lambda_2<0$

2.1.2.1)$\lambda_3=0$

From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ we obtain $\lambda_5=\lambda_0-\lambda_2>0$. I does not satisfy us.

2.1.2.2)$\lambda_3<0$

2.1.2.2.1)$\lambda_4=0$

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $0>\lambda_3=-1-1.5\lambda_2$ and from $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $0\ge\lambda_7=5+2\lambda_2+\lambda_3=4+0.5\lambda_2$. System $4+0.5\lambda_2\le 0, -1-1.5\lambda_2<0$ has no solutions.

2.1.2.2.2)$\lambda_4<0$

From $\lambda_2(3a-b+c+2d-8)=\lambda_3(2a+3b-c+d-10)=$

$=\lambda_4a=0$ we obtain $3a-b+c+2d-8=2a+3b-c+d-10=a=0$, that is $a=0, b=\frac{3}{2}(6-d), c=\frac{7}{2}\left(\frac{34}{7}-d\right)$

2.1.2.2.2.1)$d=0$. Then $b=9, c=17$.

From $\lambda_5b=\lambda_6c=0$ we obtain $\lambda_5=\lambda_6=0$.

From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ and $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_3=1>0$ and $\lambda_2=4>0$. It does not satisfy us.

2.1.2.2.2.2)$d=6$, then $b=0, c=-4$

From $\lambda_6c=\lambda_7d=0$ we obtain $\lambda_6=\lambda_7=0$. From $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_2=-\frac{2}{3}, \lambda_3=-\frac{11}{3}$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ we obtain $\lambda_4=-\frac{22}{3}, \lambda_5=-\frac{28}{3}$.

So $(0,0,-4,6)$ can be a point of global maximum.

2.1.2.2.2.3)$d=\frac{34}{7}$, then $b=\frac{12}{7}, c=0$

From $\lambda_5b=\lambda_7d=0$ we obtain $\lambda_5=\lambda_7=0$. From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_2=-2, \lambda_3=-1$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_4=-6$ and $\lambda_6=-4$.

So $\left(0,\frac{12}{7},0,\frac{34}{7}\right)$ can be a point of global maximum.

2.1.2.2.2.4)$d\not\in\left\{0,6,\frac{34}{7}\right\}$ , then $b\neq0, c\neq 0$. From $\lambda_5b=\lambda_6c=\lambda_7d=0$ we obtain $\lambda_5=\lambda_6=\lambda_7=0$. Then from $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $1-\lambda_2+3\lambda_3=0$, $-3+\lambda_2-\lambda_3=0$, $5+2\lambda_2+\lambda_3=0$. This sysem has no solutions.

2.2)$\lambda_1<0$

2.2.1)$\lambda_2=0$

2.2.1.1)$\lambda_3=0$

2.2.1.1.1)$\lambda_5=0$. From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ we obtain $\lambda_1=-\frac{1}{7}\lambda_0=-\frac{1}{7}$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $\lambda_4=2\lambda_0+\lambda_1>0$. It does not satisfy us.

2.2.1.1.2)$\lambda_6=0$. From $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_1=\lambda_0>0$. It does not satisfy us.

2.2.1.1.3)$\lambda_7=0$. From $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_1=-\frac{5}{7}\lambda_0=-\frac{5}{7}$. From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $\lambda_4=2\lambda_0+\lambda_1>0$. It does not satisfy us.

2.2.1.1.5)$0\not\in\{\lambda_5, \lambda_6, \lambda_7\}$. From $\lambda_5b=\lambda_6c=\lambda_7d=0$ we obtain $b=c=d=0$. Since $\lambda_1(a+7b+3c+7d-46)=0$ and $\lambda_1<0$, we have $a+7b+3c+7d-46=0$, that is $a=46$. Since $\lambda_4a=0$, we have $\lambda_4=0$.

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $\lambda_1=-2\lambda_0=-2$. Then from $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_5=-13, \lambda_6=-9, \lambda_7=-9$. So $(46,0,0,0)$ can be a point of global maximum.

2.2.1.2)$\lambda_3<0$

2.2.1.2.1)$\lambda_5=0$. From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ we obtain $\lambda_3=-\frac{7\lambda_1+1}{3}$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ we obtain $0\ge\lambda_4=2+\lambda_1+2\lambda_3=\frac{4-11\lambda_1}{3}$, that is $\lambda_1\ge\frac{4}{11}>0$. It does not satisfy us.

Then $\lambda_5\neq 0$. Since, in addition, $\lambda_1\neq 0$ and $\lambda_3\neq 0$, from $\lambda_1(a+7b+3c+7d-46)=$

$=\lambda_3(2a+3b-c+d-10)=\lambda_5b=0$ we obtain $a+7b+3c+7d-46=$

$=2a+3b-c+d-10=b=0$, that is $c=\frac{13}{7}\left(\frac{82}{13}-d\right), a=\frac{10}{7}\left(\frac{38}{5}-d\right)$

2.2.1.2.2)$d=0$, then $c=\frac{82}{7}, a=\frac{76}{7}$

Since $\lambda_4a=\lambda_6c=0$, we have $\lambda_4=\lambda_6=0$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_1=\frac{4}{7}>0$. It does not satisfy us.

2.2.1.2.3)$d=\frac{82}{13}$ , then $c=0, a=\frac{24}{13}$

Since $\lambda_4a=\lambda_7d=0$, we have $\lambda_4=\lambda_7=0$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_1=-\frac{4}{5}, \lambda_3=-\frac{3}{5}$. Then from $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ and $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_5=-\frac{32}{5}, \lambda_6=-\frac{24}{5}$. So $\left(\frac{24}{13},0,0,\frac{82}{13}\right)$ can be a point of global maximum.

2.2.1.2.4)$d=\frac{38}{5}$, then $c=-\frac{12}{5}, a=0$

Since $\lambda_6c=\lambda_7d=0$, we have $\lambda_6=\lambda_7=0$. Then from $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_3=-\frac{18}{5}, \lambda_1=-\frac{1}{5}$.

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ we obtain $\lambda_4=-\frac{27}{5}, \lambda_5=-\frac{56}{5}$. So $\left(0,0,-\frac{12}{5},\frac{38}{5}\right)$ can be a point of global maximum.

2.2.1.2.5)$d\not\in\left\{0,\frac{82}{13},\frac{38}{5}\right\}$, then $a\neq 0, c\neq 0$. From $\lambda_4a=\lambda_6c=\lambda_7d=0$ we obtain $\lambda_4=\lambda_6=\lambda_7=0$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain that $2+\lambda_1+2\lambda_3=0, -3+3\lambda_1-\lambda_3=0, 0=5+7\lambda_1+\lambda_3$. This has no solutions.

2.2.2)$\lambda_2<0$

2.2.2.1)$\lambda_3<0$

From $\lambda_1(a+7b+3c+7d-46)=0$, $\lambda_2(3a-b+c+2d-8)=0$, $\lambda_3(2a+3b-c+d-10)=0$ we obtain $a=-\frac{14}{33}(18+d), b=\frac{29}{66}\left(\frac{94}{29}-d\right), c=\frac{7}{6}\left(\frac{194}{7}-d\right)$

2.2.2.1.1)$d=0$, then $a=-\frac{84}{11}, b=\frac{47}{33}, c=\frac{97}{3}$

Since $\lambda_4a=\lambda_5b=\lambda_6c=0$, we have $\lambda_4=\lambda_5=\lambda_6=0$.

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_1=\frac{16}{33}>0, \lambda_2=\frac{4}{33}>0, \lambda_3=-\frac{47}{33}$ . I does not satisfy us.

2.2.2.1.2)$d=-18$, then $a=0, b=\frac{28}{3}, c=\frac{160}{3}$. Since $\lambda_5b=\lambda_6c=\lambda_7d=0$, we have $\lambda_5=\lambda_6=\lambda_7=0$.

From $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_1=-\frac{2}{7}, \lambda_2=\frac{2}{7}>0, \lambda_3=\frac{3}{7}>0$ . I does not satisfy us.

2.2.2.1.3)$d=\frac{94}{29}$, then $a=-\frac{784}{27}, b=0, c=\frac{828}{29}$

Since $\lambda_4a=\lambda_6c=\lambda_7d=0$, we have $\lambda_4=\lambda_6=\lambda_7=0$

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_1=-\frac{7}{29}, \lambda_2=\frac{4}{29}>0,\lambda_3=-\frac{63}{58}$ . It does not satisfy us.

2.2.2.1.4)$d=\frac{194}{7}$, then $a=-\frac{640}{33}, b=-\frac{828}{77}, c=0$

Since $\lambda_4a=\lambda_5b=\lambda_7d=0$, we have $\lambda_4=\lambda_5=\lambda_7=0$

From $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_1=-\frac{22}{29}, \lambda_2=\frac{54}{29}>0, \lambda_3=-\frac{99}{29}$. I does not satisfy us.

2.2.2.1.5)$d\not\in\left\{0,-18,\frac{94}{29},\frac{194}{27}\right\}$, then $a\neq 0, b\neq 0, c\neq 0$

Since $\lambda_4a=\lambda_5b=\lambda_6c=\lambda_7d=0$, we have $\lambda_4=\lambda_5=\lambda_6=\lambda_7=0$

Then $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ , $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$. This system has no solutions.

2.2.2.2)$\lambda_3=0$

2.2.2.2.1)$\lambda_6=0$. Then from $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $3\lambda_1+\lambda_2=3$. Since $\lambda_1<0, \lambda_2<0$ , it does not satisfy us.

Then $\lambda_6<0$. From $\lambda_6c=0$ we obtain $c=0$

Since $\lambda_1<0, \lambda_2<0$, and $\lambda_1(a+7b+3c+7d-46)=0$ , $\lambda_2(3a-b+c+2d-8)=0$, we obtain $a+7b+3c+7d-46=0, 3a-b+c+2d-8=0$

So $c=0, b=\frac{19}{22}\left(\frac{130}{19}-d\right), a=\frac{21}{22}\left(\frac{34}{7}-d\right)$

2.2.2.2.2)$d=0$ , then $b=\frac{65}{11}, a=\frac{51}{11}$

Since $\lambda_4a=\lambda_5b=0$, we have $\lambda_4=\lambda_5=0$. Since $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, we obtain $\lambda_1=-\frac{5}{22}, \lambda_2=-\frac{13}{22}$

Then from $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ we obtain $\lambda_7=\frac{49}{22}>0$. It does not satisfy us.

2.2.2.2.3)$d=\frac{130}{19}$, then $b=0, a=-\frac{36}{19}$

Since $\lambda_4a=\lambda_7d=0$, we have $\lambda_4=\lambda_7=0$ . Since $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$, we obtain $\lambda_1=-\frac{9}{19}, \lambda_2=-\frac{11}{19}$

Then from $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_5=-\frac{33}{19}, \lambda_6=-5$. So $\left(-\frac{36}{19},0,0,\frac{130}{19}\right)$ can be a point of global maximum.

2.2.2.2.4)$d=\frac{34}{7}$ , then $b=\frac{12}{7}, a=0$

Since $\lambda_5b=\lambda_7d=0$ , we have $\lambda_5=\lambda_7=0$. Since $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$ and $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$, we obtain $\lambda_1=-\frac{1}{3}, \lambda_2=-\frac{4}{3}$. Then from $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$ and $0=-3\lambda_0+3\lambda_1+\lambda_2-\lambda_3-\lambda_6$ we obtain $\lambda_4=-\frac{7}{3}$ and $\lambda_6=-\frac{16}{3}$. So $\left(0,\frac{12}{7},0,\frac{34}{7}\right)$ can be a point of global maximum.

2.2.2.2.5)$d\not\in\left\{0,\frac{130}{19},\frac{34}{7}\right\}$, then $a\neq 0, b\neq 0$. Since $\lambda_4a=\lambda_5b=\lambda_7d=0$, we have $\lambda_4=\lambda_5=\lambda_7=0$. Then the system $0=2\lambda_0+\lambda_1+3\lambda_2+2\lambda_3-\lambda_4$, $0=\lambda_0+7\lambda_1-\lambda_2+3\lambda_3-\lambda_5$, $0=5\lambda_0+7\lambda_1+2\lambda_2+\lambda_3-\lambda_7$ has no solutions.

Consider set of our points $A=\bigl\{(0,0,-4,6),\left(0,\frac{12}{7},0,\frac{34}{7}\right),(46,0,0,0),$

$\left(\frac{24}{13},0,0,\frac{82}{13}\right),\left(0,0,-\frac{12}{5},\frac{38}{5}\right),$

$\left(-\frac{36}{19},0,0,\frac{130}{19}\right),\left(0,\frac{12}{7},0,\frac{34}{7}\right)\bigr\}$

Let $B=\{(a,b,c,d)| a+7b+3c+7d\le 46,$

$3a-b+c+2d \le 8, 2a+3b-c+d\le 10,$

$a\ge 0, b\ge 0, c\ge 0, d\ge 0\}$.

Then $A\cap B=\left\{\left(0,\frac{12}{7},0,\frac{34}{7}\right)\right\}$

Let $f(a,b,c,d)=2a+b-3c+5d$, then $f(A\cap B)=\{26\}$.

We have that $B$ is closed as intersection of closed sets. Also $B\subset\{(a,b,c,d)|a+7b+3c+7d\le 46,$

$a\ge 0, b\ge 0, c\ge 0, d\ge 0\}\subset$

$\subset[0,46]\times\left[0,\frac{46}{7}\right]\times\left[0,\frac{46}{3}\right]\times\left[0,\frac{46}{7}\right]$. So $B$ is bounded. By the Heine–Borel theorem we obtain that $B$ is a compact set. Then since $f$ is continuous on $B$, there is a maximum of $f(B)$.

We have that the maximum of $f(B)$ is $f\left(0,\frac{12}{7},0,\frac{34}{7}\right)=26$.

Answer: $26$ in point $\left(0,\frac{12}{7},0,\frac{34}{7}\right)$