Question

Maximize Z = 3x + 4y
Subject to the constraints
x+y ≤ 4, x ≥ 0, y ≥ 0​

Answer 1

Answer:

"put x = 0 y = 0  since 0 + 0 ≤ 0, equation is true. ∴ This will be shaded towards 0.  At A (4,0), Z = 3 x 4 = 12  At B (0,4), Z = 4 x 4= 16 ∴ Z is maximised at B (0,4) = 16"

Answer 2

Answer:

$\boxed{ \bf{ \:Maximum \: value \: of \: Z = 16\: at \: (0, 4)}}$

Explanation:

Consider

$\sf \: x + y \leqslant 4$

Let we first sketch the line x + y = 4

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\qquad\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

[ See the attachment graph ]

Now, from graph we concluded that OAB is a feasible region.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: Point & \bf Value \: of \: Z = 3x + 4y\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf O(0,0) & \sf 0\\ \\ \sf A(4,0) & \sf 12\\ \\ \sf B(0, 4) & \sf 16 \end{array}} \\ \end{gathered}$

$\sf\implies \boxed{ \bf{ \:Maximum \: value \: of \: Z = 16 \: at \: (0, 4)}}$