Question
Maximize Z = 5x + 3y
Subject to
$3\text{x}+5\text{y}\leq15$
$5\text{x}+2\text{y}\leq10$
$\text{x},\text{y}\geq0$
Subject to
$3\text{x}+5\text{y}\leq15$
$5\text{x}+2\text{y}\leq10$
$\text{x},\text{y}\geq0$
✓
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 5y = 15, 5x + 2 y = 10, x = 0 and y = 0
Region represented by $3\text{x}+5\text{y}\leq15:$
The line 3x + 5y = 15 meets the coordinate axes at A(5, 0) and B(0, 3) respectively.
By joining these points we obtain the line 3x + 5y = 15.
Clearly (0, 0) satisfies the inequation $3\text{x}+5\text{y}\leq15$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+5\text{y}\leq15$.
Region represented by $5\text{x}+2\text{y}\leq10:$
The line 5x + 2y = 10 meets the coordinate axes at C(2, 0) and D(0, 5) respectively.
By joining these points we obtain the line 5x + 2y = 10.
Clearly (0, 0) satisfies the inequation $5\text{x}+2\text{y}\leq10$.
So, the region containing the origin represents the solution set of the inequation $5\text{x}+2\text{y}\leq10$.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints, $3\text{x}+5\text{y}\leq15,5\text{x}+2\text{y}\leq10,\text{x}\geq0$ and $\text{y}\geq0$,are as follows.
3x + 5y = 15, 5x + 2 y = 10, x = 0 and y = 0
Region represented by $3\text{x}+5\text{y}\leq15:$
The line 3x + 5y = 15 meets the coordinate axes at A(5, 0) and B(0, 3) respectively.
By joining these points we obtain the line 3x + 5y = 15.
Clearly (0, 0) satisfies the inequation $3\text{x}+5\text{y}\leq15$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+5\text{y}\leq15$.
Region represented by $5\text{x}+2\text{y}\leq10:$
The line 5x + 2y = 10 meets the coordinate axes at C(2, 0) and D(0, 5) respectively.
By joining these points we obtain the line 5x + 2y = 10.
Clearly (0, 0) satisfies the inequation $5\text{x}+2\text{y}\leq10$.
So, the region containing the origin represents the solution set of the inequation $5\text{x}+2\text{y}\leq10$.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints, $3\text{x}+5\text{y}\leq15,5\text{x}+2\text{y}\leq10,\text{x}\geq0$ and $\text{y}\geq0$,are as follows.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
→$\frac{\text{dy}}{\text{dx}}=\log\text{x}$
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
If $\text{y}=\sqrt{\text{x}^2+\text{a}^2},$ prvoe that $\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}=0$
→Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
→$\sqrt{\frac{1+\text{x}}{1-\text{x}}}$
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
→Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
→$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\{\log\text{y}-\log\text{x}+1\}$
Differentiate the following functions with respect to x:
$\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
→$\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}$
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
→$\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
The probability distribution of a random variable x is given as under:
$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}\text{x}^2 & \text{for}\text{ x}= 1,2,3\\2\text{kx} & \text{for}\text{ x } =4,5,6\\0&\text{otherwise} \end{cases}$
where k is a constant. Calculate
→$\text{P}(\text{X}=\text{x})=\begin{cases}\text{k}\text{x}^2 & \text{for}\text{ x}= 1,2,3\\2\text{kx} & \text{for}\text{ x } =4,5,6\\0&\text{otherwise} \end{cases}$
where k is a constant. Calculate
- $\text{E}(\text{X})$
- $\text{E}(3\text{X}^2)$
- $\text{P}(\text{X}\geq4)$
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines, and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at ₹ 7 profit and that of B at a profit of ₹ 4. Find the production level per day for maximum profit graphically.