### Linear Programming: candies

Question Sample Titled 'Linear Programming: candies'

 (a) In the figure, the equations of ${L}_{{1}}$ and ${L}_{{2}}$ are ${y}={2}{x}$ and ${x}+{2}{y}={20}$ respectively. The shaded region (including the boundary) represents the solution of a system of inequalities.

${x}$${y}$${O}$${10}$${20}$${20}$${10}$${L}_{{1}}$${L}_{{4}}$${L}_{{3}}$${L}_{{2}}$

 (i) Find the point of intersection of ${L}_{{1}}$ and ${L}_{{2}}$ . (ii) Write down the system of inequalities.  (4 marks) (b) There were candies and chocolates on sale in a candy shop. Each pack of the candies cost $3$3$ and it had ${5}$ pieces. Each pack of the chocolates cost$25$25$ and it had ${10}$ pieces. Only ${10}$ packs of candies and ${20}$ packs of chocolates were available. The number of packs of chocolates bought must be at least two times of the number of packs of candies bought. Allison bought some of the chocolates and candies from the shop as souvenirs for a festival day. On the day, Allison needs to give out at least ${100}$ pieces of sweets. Suppose Allison bought ${x}$ packs of candies and ${y}$ packs of chocolates. Allison thinks that the minimum cost of the sweets must be more than $200$200$ Do you agree? Explain your answer.  (4 marks) 題解  (ai) ${y}$ $={2}{x}$ $\ldots{\left({1}\right)}$ ${x}+{2}{y}$ $={20}$ $\ldots{\left({2}\right)}$ Sub. ${\left({1}\right)}$ into ${\left({2}\right)}$ , we have ${x}+{2}{\left({2}{x}\right)}$ $={20}$ 1M ${5}{x}$ $={20}$ ${x}$ $={4}$  ∵ ${2}{\left({4}\right)}$ $={8}$ ∴ Intersection $={\left({4},{8}\right)}$ . 1A  (aii) ${x}+{2}{y}\ge{20}$ 1M ${y}\ge{2}{x}$ ${x}\ge{0}$ 1A ${y}\le{20}$  (b) From the information, we have ${x}+{2}{y}\ge{20}$ 1M ${y}\ge{2}{x}$ ${x}\le{10}$ ${y}\le{20}$ ${x}$ , ${y}$ are non-negative integers. Cost=$(3x+25y)$=\left(3x+25y\right)$ According to the figure, test the following vertices: 2M At ${\left({0},{20}\right)}$ , cost $={3}{\left({0}\right)}+{25}{\left({20}\right)}$  =$500$=500$ At ${\left({0},{10}\right)}$ , cost $={3}{\left({0}\right)}+{25}{\left({10}\right)}$  =$250$=250$ At ${\left({4},{8}\right)}$ , cost $={3}{\left({4}\right)}+{25}{\left({8}\right)}$  =$212$=212$ At ${\left({10},{20}\right)}$ , cost $={3}{\left({10}\right)}+{25}{\left({20}\right)}$  =$530$=530$ Thus, the minimum cost is \$212$212$ . 1A ∴   She is correct.

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