Question

Solve the following problem graphically Max: Z 60x1+40X2 S t 2x1+X2≤60,X2≤25,X2≤35 and x≥0, y ≥0​

Answer 1

Answer:

$613111363 = 6161336$

$21y16x {6169133363yxy { { { { \frac{e \beta e\pi\pi\pi\% \beta \beta }{?} \times \frac{?}{?} }^{?} }^{2} }^{?} }^{2} }^{?}$

Answer 2

Answer:

max: ($Z = 60x_1+40x_2$) such that $2x_1+x_2\leq 60$, $x_2 = 25$, $x_2 = 35$ is 2050.

Step-by-step explanation:

Given,

$Z = 60x_1+40x_2\\2x_1+x_2\leq 6\\x_2\leq 25\\x_2\leq 35$

To find, max(Z)

The below graph shows the region $2x_1+x_2\leq 60$.....(1) in red colour, the region $x_2\leq 25$....(2) in blue colour, and the region $x_2\leq 35$...(3) in green colour.

The points A(17.5, 25), B(0, 25), C(0, 0), D(30, 0) are the boundaries of the common regions (1), (2), (3) because as the boundaries must be considered only in the 1st quadrant(x ≥ 0, y ≥ 0).

Now for maximisation of Z.

Then, Z(A) = 60(17.5) + 40(25) = 2050

Z(B) = 60(0) + 40(25) = 1000

Z(C) = 60(0) + 40(0) = 0

Z(D) = 60(30) + 40(0) = 1800

Therefore, max: ($Z = 60x_1+40x_2$) such that $2x_1+x_2\leq 60$, $x_2 = 25$, $x_2 = 35$ is 2050.

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