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Question-113455

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Question Number 113455 by AbhishekBasnet last updated on 13/Sep/20
Commented by som(math1967) last updated on 13/Sep/20
Number of one kind x and other kind y x≥0,y≥0 150x+75y≤1500 50x+75y≤600 Max.profit=P_(max) =5x+3y Corner points are (0,0) ,(10,0),(9,2),(0,8) ∴P_(max) at (9,2)=5×9+3×2 =Rs51
$$\mathrm{Number}\:\mathrm{of}\:\mathrm{one}\:\mathrm{kind}\:\mathrm{x}\:\mathrm{and} \\ $$$$\mathrm{other}\:\mathrm{kind}\:\mathrm{y} \\ $$$$\mathrm{x}\geqslant\mathrm{0},\mathrm{y}\geqslant\mathrm{0} \\ $$$$\mathrm{150x}+\mathrm{75y}\leqslant\mathrm{1500} \\ $$$$\mathrm{50x}+\mathrm{75y}\leqslant\mathrm{600} \\ $$$$\mathrm{Max}.\mathrm{profit}=\mathrm{P}_{\mathrm{max}} =\mathrm{5x}+\mathrm{3y} \\ $$$$\mathrm{Corner}\:\mathrm{points}\:\mathrm{are} \\ $$$$\left(\mathrm{0},\mathrm{0}\right)\:,\left(\mathrm{10},\mathrm{0}\right),\left(\mathrm{9},\mathrm{2}\right),\left(\mathrm{0},\mathrm{8}\right) \\ $$$$\therefore\mathrm{P}_{\mathrm{max}} \:\mathrm{at}\:\left(\mathrm{9},\mathrm{2}\right)=\mathrm{5}×\mathrm{9}+\mathrm{3}×\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{Rs51} \\ $$
Answered by 1549442205PVT last updated on 13/Sep/20
Suppose first kind of cake is made x boxes and second kind of cakes is made y boxes.From the hypothesis we has the system : { ((150x+75y=1500(1))),((50x+75y=600(2))) :} Substracting two equations we get 100x=900⇒x=9.Replace we get y=2.Then the profits obtain equal to 9×5+2×3=51 (Rs) •If made 10 boxes of first kind of cakes then the second kind of cakes has no boxes because 10×150=1500gr flour ,so has no flour to make second kind of cakes •From above result we infer a+b≤11 and a≤9⇒b≤11−a The profits value obtain is M=5a+3b≤5a+3(11−a)=2a+33 ≤2×9+33=51 Thus,the best option for candy product are 9 boxes of 1 type and 2 boxes of 2 type
$$\mathrm{Suppose}\:\mathrm{first}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{cake}\:\mathrm{is}\:\mathrm{made}\:\mathrm{x} \\ $$$$\mathrm{boxes}\:\mathrm{and}\:\mathrm{second}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{cakes}\:\mathrm{is} \\ $$$$\mathrm{made}\:\mathrm{y}\:\mathrm{boxes}.\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\: \\ $$$$\mathrm{we}\:\mathrm{has}\:\mathrm{the}\:\mathrm{system}\:: \\ $$$$\begin{cases}{\mathrm{150x}+\mathrm{75y}=\mathrm{1500}\left(\mathrm{1}\right)}\\{\mathrm{50x}+\mathrm{75y}=\mathrm{600}\left(\mathrm{2}\right)}\end{cases} \\ $$$$\mathrm{Substracting}\:\mathrm{two}\:\mathrm{equations}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{100x}=\mathrm{900}\Rightarrow\mathrm{x}=\mathrm{9}.\mathrm{Replace}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{y}=\mathrm{2}.\mathrm{Then}\:\mathrm{the}\:\mathrm{profits}\:\mathrm{obtain}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{9}×\mathrm{5}+\mathrm{2}×\mathrm{3}=\mathrm{51}\:\left(\mathrm{Rs}\right) \\ $$$$\bullet\mathrm{If}\:\mathrm{made}\:\mathrm{10}\:\mathrm{boxes}\:\mathrm{of}\:\mathrm{first}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{cakes} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{second}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{cakes}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{boxes}\:\mathrm{because}\:\mathrm{10}×\mathrm{150}=\mathrm{1500gr}\:\mathrm{flour} \\ $$$$,\mathrm{so}\:\mathrm{has}\:\mathrm{no}\:\mathrm{flour}\:\mathrm{to}\:\mathrm{make}\:\mathrm{second}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{cakes} \\ $$$$\bullet\mathrm{From}\:\mathrm{above}\:\mathrm{result}\:\mathrm{we}\:\mathrm{infer}\:\mathrm{a}+\mathrm{b}\leqslant\mathrm{11} \\ $$$$\mathrm{and}\:\mathrm{a}\leqslant\mathrm{9}\Rightarrow\mathrm{b}\leqslant\mathrm{11}−\mathrm{a} \\ $$$$\mathrm{The}\:\mathrm{profits}\:\mathrm{value}\:\mathrm{obtain}\:\mathrm{is} \\ $$$$\mathrm{M}=\mathrm{5a}+\mathrm{3b}\leqslant\mathrm{5a}+\mathrm{3}\left(\mathrm{11}−\mathrm{a}\right)=\mathrm{2a}+\mathrm{33} \\ $$$$\leqslant\mathrm{2}×\mathrm{9}+\mathrm{33}=\mathrm{51} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{best}\:\mathrm{option}\:\mathrm{for}\:\mathrm{candy} \\ $$$$\mathrm{product}\:\mathrm{are}\:\mathrm{9}\:\mathrm{boxes}\:\mathrm{of}\:\mathrm{1}\:\mathrm{type}\:\mathrm{and}\:\mathrm{2} \\ $$$$\mathrm{boxes}\:\mathrm{of}\:\mathrm{2}\:\mathrm{type} \\ $$

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