Inequalities and linear programming Examples


  • Examples
    (a) The figure shows ΔABC, where \(AC//DE\). Find \(AD{\rm{ :}}DB\) such that the ratio of the areas of polygon ADEC and ΔABC is \(21:25\).



    (b) (i) Draw the feasible region which represents the following constraints on a rectangular coordinate plane.
       \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\end{array} \right.\)
    (ii) Using the result of (a), if the area of the feasible region of \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\\x \le k\end{array} \right.\) is \(\frac{{21}}{{25}}\) of the area of the feasible region in (b)(i),
    find the value of k.
    (iii) Hence find the maximum value of \(f(x,{\rm{ }}y) = 3x + 2y\) subject to the constraints \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\\x \le k\end{array} \right.\).
  • Solutions

    (a) \(\frac{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f p}}{\kern 1pt} {\rm{o}}{\kern 1pt} {\rm{l}}{\kern 1pt} {\rm{y}}{\kern 1pt} {\rm{g}}{\kern 1pt} {\rm{o}}{\kern 1pt} {\rm{n }}ADEC}}{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC}} = \frac{{21}}{{25}}\)
    \(\frac{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC - {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta DBE}}{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC}} = \frac{{21}}{{25}}\)
    \(\begin{array}{1}25 \times {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC - 25 \times {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta DBE = 21 \times {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC\\4 \times {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC = 25 \times {\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta DB{E^{^{^{}}}}\\\frac{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta DBE}}{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC}} = {\frac{4}{{25}}^{^{}}}\end{array}\)
    ∵ \(\Delta DBE\~\Delta ABC\) (equiangular)
    ∴ \(\frac{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta DBE}}{{{\rm{A}}{\kern 1pt} {\rm{r}}{\kern 1pt} {\rm{e}}{\kern 1pt} {\rm{a o}}{\kern 1pt} {\rm{f }}\Delta ABC}} = {(\frac{{DB}}{{AB}})^2}\)
    \(\begin{array}{1}\frac{4}{{25}} = {(\frac{{DB}}{{AD + DB}})^2}\\\frac{2}{5} = {\frac{{DB}}{{AD + DB}}^{^{}}}\\2AD + 2DB = 5D{B^{^{^{}}}}\\2AD = 3D{B^{^{}}}\\\frac{{AD}}{{DB}} = {\frac{3}{2}^{}}\end{array}\)
    ∴ Ratio required \( = \underline{\underline {3\;{\rm{:}}2}} \)

    (b) (i) [graphMissing Inequalities and linear programming 02 Q56]



    The shaded region in the graph is the feasible region.
    (ii) [graphMissing Inequalities and linear programming 02 Q56]


    The shaded region in the graph is the feasible region of \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\\x \le k\end{array} \right.\).
    Using the result of (a), if the area of the feasible region of \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\\x \le k\end{array} \right.\) is \(\frac{{21}}{{25}}\) of the area of the feasible region in (b)(i), then \(AD\;{\rm{:}}DB = 3\;{\rm{:}}2\).
    ∴ \(\begin{array}{1}k = \frac{{3(5) + 2(0)}}{{3 + 2}}\\ = \underline{\underline 3} \end{array}\)
    (iii) The feasible solutions of the constraints \(\left\{ \begin{array}{l}2x - 3y \le 7\\x + y \le 6\\x \ge 0\\x \le k\end{array} \right.\) are represented graphically as below.
    [graphMissing Inequalities and linear programming 02 Q56]
    Draw the straight line \(f(x,{\rm{ }}y) = 0\).
    From the graph, \(f(x,{\rm{ }}y)\) attains its maximum / minimum values at the points \((0,{\rm{ }} - \frac{7}{3})\) and \((3,{\rm{ }}3)\).
    At the point \((0,{\rm{ }} - \frac{7}{3})\), \(f(0,{\rm{ }} - \frac{7}{3}) = 3(0) + 2( - \frac{7}{3}) = - \frac{{14}}{3}\).
    At the point \((3,{\rm{ }}3)\), \(f(3,{\rm{ }}3) = 3(3) + 2(3) = 15\).
    ∴ Maximum value\( = \underline{\underline {15}} \)


  • Examples
    The table below shows the potassium, phosphorus and nitrogen contents in each kg of fertilizer A and each kg of fertilizer B.
    [graphMissing Inequalities and linear programming 02 Q55]
    A merchant mixed fertilizer A with fertilizer B such that each package of mixture contains at least 12 units of potassium, 10 units of phosphorus and 8 units of nitrogen. Suppose there are x kg of fertilizer A and y kg of fertilizer B in each package of mixture,
    (a) write down all the constraints about x and y.
    (b) represent the feasible solutions on a rectangular coordinate plane.
    (c) Given that the costs of each kg of fertilizer A and each kg of fertilizer B are $9 and $6 respectively, find the minimum cost of each package of mixture.
  • Solutions

    (a) The constraints are
        \(\left\{ \begin{array}{l}6x + 2y \ge 12\\2x + 4y \ge 10\\2x + 2y \ge 8\\x \ge 0\\y \ge 0\end{array} \right.\),
    which are equivalent to
        \(\left\{ \begin{array}{l}3x + y \ge 6\\x + 2y \ge 5\\x + y \ge 4\\x \ge 0\\y \ge 0\end{array} \right.\).

    (b) [graphMissing Inequalities and linear programming 02 Q55]
    The shaded region in the figure represents the feasible solutions.

    (c) Cost of each package of mixture \(\begin{array}{1}\$ C(x,{\rm{ }}y) = \$ (9x + 6y)\\ = \$ 3(3x + 2y)\end{array}\)
    Draw the straight line \(3(3x + 2y) = 0\), i.e. the straight line \(3x + 2y = 0\) on the graph in (b).
    From the graph, the cost of each package of mixture is the minimum when \(x = 1\) and \(y = 3\).
    ∴ Minimum cost of each package of mixture\(\begin{array}{l} = \$ 3(3x + 2y)\\ = \$ 3(3 \times 1 + 2 \times 3)\\ = \underline{\underline {\$ 27}} \end{array}\)


  • Examples
    A merchant buys 450 kg of brand A rice, 325 kg of brand B rice and 500 kg of brand C rice. He produces x kg of mixture X and y kg of mixture Y by mixing brands A, B and C rice in the ratios of 3 : 2 : 1 and 1 : 1 : 2 respectively.
    (a) Write down all the constraints about x and y.
    (b) Represent the feasible solutions on a rectangular coordinate plane.
    (c) It is given that the profit of selling 1 kg of mixture X is 1.5 times the profit of selling 1 kg of mixture Y.
        (i) How many kg of each mixture should be produced to obtain the maximum profit?
        (ii) When the profit is the maximum, how many kg of each brand of rice are used to produce the mixtures?
  • Solutions

    (a) The constraints are
    \(\left\{ {\begin{array}{*{20}{l}}{\frac{3}{{3 + 2 + 1}}x + \frac{1}{{1 + 1 + 2}}y \le 450}\\{\frac{2}{{3 + 2 + 1}}x + \frac{1}{{1 + 1 + 2}}y \le {{325}^{^{^{^{^{}}}}}}}\\{\frac{1}{{3 + 2 + 1}}x + \frac{2}{{1 + 1 + 2}}y \le {{500}^{^{^{^{^{}}}}}}}\\{x \ge {0^{^{^{}}}}}\\{y \ge 0}\end{array}} \right.\),
    which are equivalent to
    \(\left\{ \begin{array}{l}2x + y \le 1{\rm{ }}800\\4x + 3y \le 3{\rm{ }}900\\x + 3y \le 3{\rm{ }}000\\x \ge 0\\y \ge 0\end{array} \right.\).

    (b) [graphMissing Inequalities and linear programming 02 Q58]
    The shaded region in the figure represents all feasible solutions.

    (c) (i) Suppose the profit of selling 1 kg of mixture Y is $k (where \(k > 0\)),
        then the profit of selling 1 kg of mixture X is $1.5k.
        Total profit \(\begin{array}{c}\$ P(x,{\rm{ }}y) = \$ (1.5kx + ky)\\ = \$ \frac{k}{2}(3x + 2y)\end{array}\)
        Draw the straight line \(\frac{k}{2}(3x + 2y) = 0\), i.e. the straight line \(3x + 2y = 0\) on the graph in (b).
        From the graph, the total profit is the maximum when \(x = 750\) and \(y = 300\).
        ∴ 750 kg of mixture X and 300 kg of mixture Y should be produced.
        (ii) When the profit is the maximum,
        weight of brand A rice used\(\begin{array}{l} = (750 \times \frac{3}{{3 + 2 + 1}} + 300 \times \frac{1}{{1 + 1 + 2}}){\rm{ kg}}\\ = \underline{\underline {450{\rm{ kg}}}} \end{array}\)
        weight of brand B rice used\(\begin{array}{l} = (750 \times \frac{2}{{3 + 2 + 1}} + 300 \times \frac{1}{{1 + 1 + 2}}){\rm{ kg}}\\ = \underline{\underline {325{\rm{ kg}}}} \end{array}\)
        weight of brand C rice used\(\begin{array}{l} = (750 \times \frac{1}{{3 + 2 + 1}} + 300 \times \frac{2}{{1 + 1 + 2}}){\rm{ kg}}\\ = \underline{\underline {275{\rm{ kg}}}} \end{array}\)