The maximum value of 2x+y subject to 3x+5y≤26 and 5x+3y≤30, x≥0, y≥0 is

MHT CET · Maths · Area Under Curves

The maximum value of $2 x + y$ subject to $3 x + 5 y \leq 26$ and $5 x + 3 y \leq 30, x \geq 0, y \geq 0$ is

A $12$
B $11.5$
C $10$
D $17.33$

Verified Solution

Answer & Solution

Correct Answer

(A) $12$

Step-by-step Solution

Detailed explanation

3 x + 5 y = 26.... (i) \times 5

5 x + 3 y = 30.... (i i) \times 3

15 x + 25 y = 130

15 x + 9 y = 90

16 y = 40

y = \frac{40}{16} = \frac{5}{2}

∴ 3 x + \frac{5 \times 5}{2} = 26

3 x = 26 - \frac{25}{2}

x = \frac{9}{2}

z = 2 x + y

Now check values of objective function at corner points of the shaded region

Z_{A} = 2 \times 6 + 0 = 12

Z_{B} = 2 \times \frac{9}{2} + \frac{5}{2} = 9 + 2.5 = 11.5

Z_{C} = 2 \times 0 + \frac{26}{5} = \frac{26}{5} = 5.2

Hence

Z_{\max} = 12

x = 6

and

y = 0

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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