Question Number 187595 by Ari last updated on 19/Feb/23
Commented by Ari last updated on 19/Feb/23
Who can help me with problem please!I can not create a function in this situation
Answered by mr W last updated on 19/Feb/23
$${x}={number}\:{of}\:{tables}\:{of}\:{type}\:\mathrm{1} \\ $$$${y}={number}\:{of}\:{tables}\:{of}\:{type}\:\mathrm{2} \\ $$$${c}={number}\:{of}\:{customers} \\ $$$$\left.{a}\right) \\ $$$${c}=\mathrm{4}{x}+\mathrm{6}{y} \\ $$$${this}\:{is}\:{a}\:{function}\:{between}\:{x},\:{y}\:{and}\:{c}. \\ $$$$\left.{b}\right) \\ $$$${c}=\mathrm{4}{x}+\mathrm{6}{y}=\mathrm{50} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{25} \\ $$$$\Rightarrow\mathrm{2}{x}=\mathrm{25}−\mathrm{3}{y} \\ $$$${we}\:{see}\:{y}\:{must}\:{be}\:{odd}\:{such}\:{that}\:\mathrm{25}−\mathrm{3}{y} \\ $$$${is}\:{even}. \\ $$$$\mathrm{0}\leqslant\mathrm{25}−\mathrm{3}{y}\: \\ $$$$\Rightarrow{y}\leqslant\mathrm{8}\:\Rightarrow{y}=\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}. \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{4}\:{possibilities}: \\ $$$$\mathrm{1}\:{table}\:{of}\:{type}\:\mathrm{2}\:{and}\:\mathrm{11}\:{tables}\:{of}\:{type}\:\mathrm{1}, \\ $$$$\mathrm{3}\:{tables}\:{of}\:{type}\:\mathrm{2}\:{and}\:\mathrm{8}\:{tables}\:{of}\:{type}\:\mathrm{1}, \\ $$$$\mathrm{5}\:{tables}\:{of}\:{type}\:\mathrm{2}\:{and}\:\mathrm{5}\:{tables}\:{of}\:{type}\:\mathrm{1}, \\ $$$$\mathrm{7}\:{tables}\:{of}\:{type}\:\mathrm{2}\:{and}\:\mathrm{2}\:{tables}\:{of}\:{type}\:\mathrm{1}. \\ $$
Commented by Ari last updated on 19/Feb/23
Thank you very much sir!