Maximize Z = 15x + 10y

Subject to

3x+2y≤802x+3y≤70  x, y≥0

Asked by Sakshi | 1 year ago |  194

##### Solution :-

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y = 0

Region represented by 3x + 2y ≤ 80 :
The line 3x + 2y = 80 meets the coordinate axes at A($$\dfrac{80}{3}$$, 0) and B(0, 40) respectively. By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation 3x + 2y ≤ 80 . So,the region containing the origin represents the solution set of the inequation 3x + 2y ≤ 80 .

Region represented by 2x + 3y ≤ 70:
The line 2x + 3y = 70 meets the coordinate axes at C(35, 0) and D(0,$$\dfrac{70}{3}$$) respectively. By joining these points we obtain the line 2x + 3y ≤ 70.
Clearly (0,0) satisfies the inequation 2x + 3y ≤ 70. So,the region containing the origin represents the solution set of the inequation 2x + 3y ≤ 70.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints, 3x + 2y ≤ 80, 2x + 3y ≤ 70, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are O(0, 0), A(803, 0), E(20, 10) and D(0,$$\dfrac{70}{3}$$)

The values of Z at these corner points are as follows.

 Corner point Z = 15x + 10y O(0, 0) 15 × 0 + 10 × 0 = 0 $$A(\dfrac{80}{3}, 0)$$ $$15 × \dfrac{80}{3} + 10 × 0 = 400$$ E(20, 10) 15 × 20 + 10 × 10 = 400 $$D(0,\dfrac{70}{3})$$ $$15 × 0 + 10 ×\dfrac{70}{3} = \dfrac{700}{3}$$

We see that the maximum value of the objective function Z is 400 which is at A($$\dfrac{80}{3}$$, 0) and E(20, 10)
Thus, the optimal value of Z is 400.

Answered by Aaryan | 1 year ago

### Related Questions

#### Minimize Z = 2x + 4y

Minimize Z = 2x + 4y

Subject to

x+y≥8

x+4y≥12

x≥3, y≥2

#### Maximize Z = 7x + 10y

Maximize Z = 7x + 10y

Subject to

x+y≤30000

y≤12000

x≥6000

x≥y

x, y≥0

#### Maximize Z = 3x + 4y

Maximize Z = 3x + 4y

Subject to

2x+2y≤802x+4y≤120

#### Maximize Z = 10x + 6y

Maximize Z = 10x + 6y

Subject to

3x+y≤122x+5y≤34  x, y≥0