Question Number 156584 by Ghaniy last updated on 12/Oct/21
$$\mathrm{find}\:\mathrm{p}\:\mathrm{if}\:\mathrm{y}=\mathrm{1}−\mathrm{px}−\mathrm{3x}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{13}\:\:\left(\mathrm{help}\:\mathrm{pls}\right) \\ $$
Commented by john_santu last updated on 14/Oct/21
$${max}\:{f}\left({x}\right)=−\mathrm{3}{x}^{\mathrm{2}} −{px}+\mathrm{1} \\ $$$${when}\:{x}=−\left(\frac{−{p}}{\mathrm{2}\left(−\mathrm{3}\right)}\right)=−\frac{{p}}{\mathrm{6}} \\ $$$${so}\:\Rightarrow\mathrm{13}=\mathrm{1}−{p}\left(−\frac{{p}}{\mathrm{6}}\right)−\mathrm{3}\left(\frac{{p}^{\mathrm{2}} }{\mathrm{36}}\right) \\ $$$$\Rightarrow\mathrm{12}=\frac{{p}^{\mathrm{2}} }{\mathrm{6}}−\frac{{p}^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{12}=\frac{{p}^{\mathrm{2}} }{\mathrm{12}}\Rightarrow{p}=\pm\mathrm{12} \\ $$
Answered by mr W last updated on 13/Oct/21
$${y}=\mathrm{1}−{px}−\mathrm{3}{x}^{\mathrm{2}} \\ $$$${y}=\mathrm{1}−\mathrm{3}\left(\frac{{p}}{\mathrm{3}}{x}+{x}^{\mathrm{2}} \right) \\ $$$${y}=\mathrm{1}−\mathrm{3}\left(\mathrm{2}\frac{{p}}{\mathrm{6}}{x}+{x}^{\mathrm{2}} \right) \\ $$$${y}=\mathrm{1}+\frac{{p}^{\mathrm{2}} }{\mathrm{12}}−\mathrm{3}\left(\frac{{p}^{\mathrm{2}} }{\mathrm{36}}+\mathrm{2}\frac{{p}}{\mathrm{6}}{x}+{x}^{\mathrm{2}} \right) \\ $$$${y}=\mathrm{1}+\frac{{p}^{\mathrm{2}} }{\mathrm{12}}−\mathrm{3}\left({x}+\frac{{p}}{\mathrm{6}}\right)^{\mathrm{2}} \leqslant\mathrm{1}+\frac{{p}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${y}_{{max}} =\mathrm{1}+\frac{{p}^{\mathrm{2}} }{\mathrm{12}}=\mathrm{13} \\ $$$${p}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} \\ $$$$\Rightarrow{p}=\pm\mathrm{12} \\ $$
Commented by otchereabdullai@gmail.com last updated on 13/Oct/21
$$\mathrm{nice}\:\mathrm{question}+\:\mathrm{nice}\:\mathrm{solution} \\ $$
Commented by Ghaniy last updated on 14/Oct/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{Mr}\:\mathrm{W} \\ $$