Question Number 33136 by Rio Mike last updated on 11/Apr/18
$${given}\:{that}\:{y}=\:\mathrm{3}{x}^{\mathrm{4}} \:{find}\:{the}\:{percentage} \\ $$$${increase}\:{in}\:{y}\:{when}\:{x}\:{increases}\:{at}\:\frac{\mathrm{5}}{\mathrm{2}}\%. \\ $$$$ \\ $$
Answered by MJS last updated on 11/Apr/18
$${x}+\frac{\mathrm{5}}{\mathrm{2}}\%={x}×\left(\mathrm{1}+\frac{\mathrm{5}/\mathrm{2}}{\mathrm{100}}\right)=\frac{\mathrm{41}}{\mathrm{40}}{x} \\ $$$$\left(\frac{\mathrm{41}}{\mathrm{40}}{x}\right)^{\mathrm{4}} =\frac{\mathrm{2825761}}{\mathrm{2560000}}{x}^{\mathrm{4}} = \\ $$$$={y}×\left(\mathrm{1}+\frac{\mathrm{2825761}/\mathrm{25600}}{\mathrm{100}}\right)= \\ $$$$={y}+\frac{\mathrm{265761}}{\mathrm{25600}}\%\approx{y}+\mathrm{10}.\mathrm{38\%} \\ $$
Commented by Rio Mike last updated on 11/Apr/18
$${sir}\:{do}\:{we}\:{get}\:{y}+\mathrm{231\%}\:{at}\:{such}\:{state} \\ $$$${my}\:{result}\:{seems}\:{to}\:{be}\:{different}\:\mathrm{10}.\mathrm{38\%} \\ $$
Commented by MJS last updated on 11/Apr/18
$$\mathrm{you}'\mathrm{re}\:\mathrm{right},\:\mathrm{I}\:\mathrm{forgot}\:\mathrm{the}\:\mathrm{factor}\:\mathrm{3}... \\ $$$$\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$