X1 x2 and x3 that maximize fx1 x2 x3 3x1 4x2 6x3 subject to...

$x_1$, $x_2$, and $x_3$ that maximize $f(x_1, x_2, x_3) = 3x_1 + 4x_2 + 6x_3$ subject to the following constraints. Write the exact answer. Do not round. If the function has no maximum value, write None. If infinitely many values maximize the function, write Inf. $$ \begin{cases} 2x_1 + x_2 + x_3 \le 11 \\ x_1 + 3x_2 + x_3 \le 17 \\ x_1 + x_2 + 4x_3 \le 18 \end{cases} $$ Answer Tables Keypad Keyboard Shortcuts $x_1 = $ $x_2 = $ Submit Answer © 2025 Hawkes Learning

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00:01 Hi, from the question given that so maximum z is x1 plus 2 x2 plus 3 x3 plus x4 plus x5 minus 2 x3 and subject to the contraign is x 1 plus 2 x2 plus 2 x3 plus x 4 plus x5 plus x5 less than are equal to 12 x1 plus 2x3 plus x5 plus x6 less than are equal to 80 3x1 plus xx2 plus x2 plus x5 less than are equal to 24 here for all x i is greater than are equal to zero here the problem is converted to a canonical form by adding this slack or surplus or artificial variable so here as the constraint negative one is of type less than or equal to that is constraint one this is constraint one is of type less than are equal to therefore we should add slack variable so second is second equation also we should add slack variables since the constraints are less than are equal to third one is also we have to add the slack variable s 3 since the constraints are less than are equal to so here add the slack variables for the constraint 1 is s 1 for the constraint 2 s 2 for the constraint 3 s 3 so here x i s i are greater than are equal to 0 now here first enter the slack variable b is s 1 s 2 s 3 and the c b is 0 c j is nothing but the coefficient of the maximum z.

01:43 So, cb is 0 ,000 and xb is the values that are present in the left side of the equation.

01:55 So 12, 18, 24.

01:57 So here 12, 18, 24 and x1, coefficient of x1, x2, x3, x4, x5, x6 and s1, s2, s3.

02:05 Now, here first we need to find z is equal to 0, then zj is also 0.

02:10 So first we need to find zj minus cj.

02:14 So zj minus cj minus zj is 0 minus 1 minus 1.

02:17 So here 0 minus 2 is minus 2.

02:20 0 minus 3 is minus 3.

02:22 So similarly we will get like this that is minus 1 minus 2 0 0.

02:26 Here we need to find from zj minus zj we need to find the minimum value.

02:31 So here the least value is negative 3.

02:35 So negative 3 is present in the column x3.