Worksheet · Grade 12 · Applied Mathematics
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6 questions · 6 marks
The graph of the inequality \(3x + 2y \le 12\) is:
Answer: (b) the half-plane on the side of the line \(3x + 2y = 12\) that contains the origin, together with the points on the line
Testing the origin: \(3(0) + 2(0) = 0 \le 12\) is true, so the origin's side is shaded; the sign \(\le\) includes the boundary line itself.
Any point in the feasible region of a linear programming problem is called a ______ solution of the problem.
Answer: feasible
Every point of the feasible region satisfies all the constraints, including the non-negative ones, so it is a feasible solution; a point outside it is an infeasible solution.
A feasible region that can be enclosed within a circle is called a bounded region.
Answer: True
This is the meaning of a bounded feasible region; a feasible region that extends without end in some direction and cannot be enclosed within any circle is unbounded.
What is an optimal feasible solution of a linear programming problem?
Answer: It is a point of the feasible region that gives the optimal value (the maximum or the minimum, as required) of the objective function.
The corner points of the feasible region given by \(x + y \le 5\), \(x \le 3\), \(x \ge 0\), \(y \ge 0\) are:
Answer: (c) \((0, 0)\), \((3, 0)\), \((3, 2)\), \((0, 5)\)
The line \(x = 3\) cuts \(x + y = 5\) at \((3, 2)\), so the region is the quadrilateral with corners \((0, 0)\), \((3, 0)\), \((3, 2)\) and \((0, 5)\); the point \((3, 5)\) breaks \(x + y \le 5\).
The constraints of an LPP are \(x + 2y \le 10\), \(2x + y \le 6\), \(x \ge 0\), \(y \ge 0\). Is \((2, 3)\) a feasible or an infeasible solution? Give a reason.
Answer: Infeasible: \(x + 2y = 8 \le 10\) holds, but \(2x + y = 7 > 6\), so the point breaks the second constraint.
Build on it
8 questions · 11 marks
Maximise \(Z = 2x + 5y\) subject to \(x + y \le 6\), \(y \le 4\), \(x \ge 0\), \(y \ge 0\). The maximum value of \(Z\) is:
Answer: (c) 24
Corner points: \((0, 0)\), \((6, 0)\), \((2, 4)\), \((0, 4)\), giving \(Z = 0, 12, 24, 20\). The maximum is 24 at \((2, 4)\); 30 would need \((0, 6)\), which breaks \(y \le 4\).
Assertion (A): The feasible region given by \(x + y \ge 4\), \(x \ge 0\), \(y \ge 0\) is unbounded.
Reason (R): A feasible region is unbounded whenever it lies in the first quadrant.
Answer: (c) A is true, but R is false
The region on or above \(x + y = 4\) in the first quadrant goes on without end, so it cannot be enclosed in a circle: A is true. R is false: the triangle given by \(x + y \le 4\), \(x \ge 0\), \(y \ge 0\) also lies in the first quadrant and is bounded.
If no point satisfies all the constraints of an LPP together, the problem has no feasible region and is said to be ______.
Answer: infeasible
When the half-planes of the constraints have no common region, there is no feasible solution at all, so the LPP is infeasible and has no optimal solution.
The feasible region of an LPP is given by \(3x + y \le 15\), \(x + 4y \le 16\), \(x \ge 0\), \(y \ge 0\). Its corner points are \(O(0, 0)\), \(A(5, 0)\), \(B(4, 3)\) and \(C(0, 4)\). Match each optimum in Column A with the corner point in Column B where it occurs.
| Column A | Column B |
|---|---|
| (i) Maximum of \(Z = x\) | (P) \(O(0, 0)\) |
| (ii) Maximum of \(Z = y\) | (Q) \(A(5, 0)\) |
| (iii) Maximum of \(Z = x + y\) | (R) \(B(4, 3)\) |
| (iv) Minimum of \(Z = x + y\) | (S) \(C(0, 4)\) |
Answer: (i)-Q, (ii)-S, (iii)-R, (iv)-P
At O, A, B, C: \(x = 0, 5, 4, 0\) (largest at A); \(y = 0, 0, 3, 4\) (largest at C); \(x + y = 0, 5, 7, 4\) (largest at B, smallest at O).
Solve graphically: Maximise \(Z = 4x + 3y\) subject to \(x + y \le 8\), \(x \le 5\), \(y \le 6\), \(x \ge 0\), \(y \ge 0\).
Answer: The feasible region is the bounded pentagon in the first quadrant below \(x + y = 8\), to the left of \(x = 5\) and below \(y = 6\).
Corner points: \((0, 0)\), \((5, 0)\), \((5, 3)\) (where \(x = 5\) meets \(x + y = 8\)), \((2, 6)\) (where \(y = 6\) meets \(x + y = 8\)) and \((0, 6)\).
Values of \(Z\): \(0\), \(20\), \(4(5) + 3(3) = 29\), \(4(2) + 3(6) = 26\), \(18\).
Maximum \(Z = 29\) at \(x = 5\), \(y = 3\).
How marks are given
Is the feasible region given by \(x \ge 2\), \(y \ge 1\), \(x + y \le 6\) bounded or unbounded? Write its corner points.
Answer: Bounded: it is the triangle with corner points \((2, 1)\), \((5, 1)\) and \((2, 4)\).
The feasible region given by \(x + 2y \ge 10\), \(2x + y \ge 8\), \(x \ge 0\), \(y \ge 0\) is unbounded, with corner points \((0, 8)\), \((2, 4)\) and \((10, 0)\). For \(Z = 2x + 3y\) on this region:
Answer: (c) the minimum is 16, at \((2, 4)\), and there is no maximum
At the corners \(Z = 24, 16, 20\). The open half-plane \(2x + 3y < 16\) has no point in common with the region, so 16 at \((2, 4)\) is the minimum; \(Z\) grows without limit on the region, so there is no maximum. \((0, 0)\) is not in the region.
Assertion (A): If \(Z = 2x + y\) takes its maximum value on a bounded feasible region at two corner points, the LPP has infinitely many optimal solutions.
Reason (R): On a bounded feasible region, the objective function always has both a maximum and a minimum value.
Answer: (b) Both A and R are true, but R is NOT the correct explanation for A
A is true: every point on the segment joining the two corners gives the same maximum value. R is also true, but it only says an optimum exists; it does not explain why there are many optimal points.
Stretch yourself
6 questions · 24 marks
The feasible region of an LPP is given by \(x + y \ge 6\), \(x \le 4\), \(x \ge 0\), \(y \ge 0\).
(a) Find its corner points and say whether it is bounded.
(b) Find the minimum value of \(Z = 3x + 2y\), checking that it really is the minimum.
(c) Does \(Z\) have a maximum on this region? Give a reason.
Answer: (a) Corner points: \((0, 6)\) and \((4, 2)\) (where \(x = 4\) meets \(x + y = 6\)). The region goes upwards without end, so it is unbounded.
(b) \(Z(0, 6) = 12\) and \(Z(4, 2) = 16\), so the smallest corner value is 12. As the region is unbounded, check the open half-plane \(3x + 2y < 12\): every point of the region has \(y \ge 6 - x\), so \(3x + 2y \ge 3x + 12 - 2x = x + 12 \ge 12\). No point of the region gives less than 12, so the minimum is 12 at \((0, 6)\).
(c) No. Along \((0, 6 + k)\), \(Z = 12 + 2k\), which grows without limit as \(k\) increases, so \(Z\) has no maximum.
How marks are given
The feasible region of an LPP is the quadrilateral with corner points \(O(0, 0)\), \(A(7, 0)\), \(B(5, 4)\) and \(C(0, 6)\). Find the system of linear inequalities whose feasible region this is.
Answer: Line AB through \((7, 0)\) and \((5, 4)\): slope \(= \dfrac{4 - 0}{5 - 7} = -2\), so \(y = -2(x - 7)\), i.e. \(2x + y = 14\).
Line BC through \((5, 4)\) and \((0, 6)\): slope \(= \dfrac{6 - 4}{0 - 5} = -\dfrac{2}{5}\), so \(y = 6 - \dfrac{2}{5}x\), i.e. \(2x + 5y = 30\).
The origin lies in the region and gives \(0 < 14\) and \(0 < 30\), so both are of the type \(\le\).
System: \(2x + y \le 14\), \(2x + 5y \le 30\), \(x \ge 0\), \(y \ge 0\).
How marks are given
A school needs at least 160 seats for a trip. A large bus has 40 seats and costs ₹5,000 to hire; a minibus has 20 seats and costs ₹3,000. Only 3 large buses and 6 minibuses are available. Using the corner point method, find how many of each the school should hire for the least cost.
Answer: Let \(x\) large buses and \(y\) minibuses be hired.
Minimise \(Z = 5000x + 3000y\) subject to \(40x + 20y \ge 160\) (i.e. \(2x + y \ge 8\)), \(x \le 3\), \(y \le 6\), \(x \ge 0\), \(y \ge 0\).
The feasible region is the triangle with corner points \((1, 6)\), \((3, 6)\) and \((3, 2)\).
\(Z(1, 6) = 23000\), \(Z(3, 6) = 33000\), \(Z(3, 2) = 21000\).
Least cost ₹21,000: hire 3 large buses and 2 minibuses.
How marks are given
A workshop makes standard and deluxe school bags. A standard bag needs 2 hours of cutting and 2 hours of stitching; a deluxe bag needs 3 hours of cutting and 1 hour of stitching. Each week at most 60 hours of cutting and 40 hours of stitching are available, and the shop can sell at most 16 deluxe bags. The profit is ₹150 on a standard bag and ₹200 on a deluxe bag. Formulate the LPP and solve it by the corner point method. At the best plan, which of the three limits are fully used?
Answer: Let \(x\) standard and \(y\) deluxe bags be made a week.
Maximise \(Z = 150x + 200y\) subject to \(2x + 3y \le 60\) (cutting), \(2x + y \le 40\) (stitching), \(y \le 16\) (sales), \(x \ge 0\), \(y \ge 0\).
Corner points of the bounded feasible region:
\(O(0, 0)\); \(A(20, 0)\) on \(2x + y = 40\);
\(B(15, 10)\): \(2x + 3y = 60\) and \(2x + y = 40\) give \(2y = 20\), so \(y = 10\), \(x = 15\);
\(C(6, 16)\): \(y = 16\) in \(2x + 3y = 60\) gives \(x = 6\);
\(D(0, 16)\).
Values: \(Z(O) = 0\), \(Z(A) = 3000\), \(Z(B) = 2250 + 2000 = 4250\), \(Z(C) = 900 + 3200 = 4100\), \(Z(D) = 3200\).
Greatest profit ₹4,250 with 15 standard and 10 deluxe bags.
At this plan cutting uses \(30 + 30 = 60\) hours and stitching \(30 + 10 = 40\) hours, so both are fully used; only 10 of the 16 deluxe bags that can be sold are made, so the sales limit is not reached.
How marks are given
A gardener mixes two fertilisers, P and Q. Each kg of P contains 3 units of nitrogen, 1 unit of phosphate and 1 unit of potash; each kg of Q contains 1 unit of nitrogen, 2 units of phosphate and 1 unit of potash. The garden needs at least 9 units of nitrogen, 8 units of phosphate and 6 units of potash. P costs ₹30 per kg and Q costs ₹40 per kg. Formulate the LPP and find the cheapest mix graphically. The feasible region is unbounded: show that the minimum you find is truly the least cost.
Answer: Let the mix contain \(x\) kg of P and \(y\) kg of Q.
Minimise \(Z = 30x + 40y\) subject to \(3x + y \ge 9\), \(x + 2y \ge 8\), \(x + y \ge 6\), \(x \ge 0\), \(y \ge 0\).
Corner points of the (unbounded) feasible region:
\((0, 9)\) on the \(y\)-axis;
\((1.5, 4.5)\), where \(3x + y = 9\) meets \(x + y = 6\);
\((4, 2)\), where \(x + y = 6\) meets \(x + 2y = 8\);
\((8, 0)\) on the \(x\)-axis.
Values: \(Z(0, 9) = 360\), \(Z(1.5, 4.5) = 45 + 180 = 225\), \(Z(4, 2) = 120 + 80 = 200\), \(Z(8, 0) = 240\).
The smallest corner value is 200. Check the open half-plane \(30x + 40y < 200\), i.e. \(3x + 4y < 20\): for every feasible point, \(3x + 4y = 2(x + y) + (x + 2y) \ge 2(6) + 8 = 20\), so no feasible point lies in it.
Hence the least cost is ₹200, with 4 kg of P and 2 kg of Q.
How marks are given
A caterer is preparing lunch for a large wedding and offers two kinds of plates: a veg thali and a special thali. The hall can seat at most 300 guests, so at most 300 plates can be served in all. Each veg thali uses 200 g of rice and each special thali uses 100 g of rice, and the caterer has 50 kg of rice. Each special thali also needs 100 g of paneer, and only 20 kg of paneer can be delivered in time. The caterer earns a profit of ₹40 on a veg thali and ₹60 on a special thali. Taking \(x\) veg thalis and \(y\) special thalis, she writes:
Maximise \(Z = 40x + 60y\)
subject to \(x + y \le 300\), \(2x + y \le 500\), \(y \le 200\), \(x \ge 0\), \(y \ge 0\).
She now has to choose the plan that gives her the greatest profit, and she also wants to know whether ordering more paneer would be worth it.
(i) No: \(150 + 160 = 310 > 300\), so it breaks the seating constraint \(x + y \le 300\).
(ii) Subtracting, \(x = 200\), and then \(y = 100\): the point is \((200, 100)\).
(iii) Corner points \((0, 0)\), \((250, 0)\), \((200, 100)\), \((100, 200)\), \((0, 200)\) give \(Z = 0, 10000, 14000, 16000, 12000\). Greatest profit ₹16,000 with 100 veg and 200 special thalis.
(iv) The new corner points are \((0, 0)\), \((250, 0)\), \((200, 100)\) and \((0, 300)\), giving \(Z = 0, 10000, 14000, 18000\). The best plan is now 300 special thalis and no veg thalis, for a profit of ₹18,000, which is ₹2,000 more than before.
