Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/3given \( \begin{align*} \mathrm{{f{{\left({x},{y}\right)}}}} &= \mathrm{{x}^{{2}}+{y}^{{2}}} \end{align*} \)at which , \( \begin{align*} \mathrm{{0}} &\le \mathrm{{x},{y}} \end{align*} \)and \( \begin{align*} \mathrm{{3}{x}+{9}{y}} &\le \mathrm{{8}} \end{align*} \)first obtaining critical points of function by putting :\( \mathrm{{{f}_{{{x}}}}} \) = 0 and \( \mathrm{{{f}_{{{y}}}}} \) = 0 \( \mathrm{{{f}_{{{x}}}}} \) = 2x , \( \mathrm{{{f}_{{{y}}}}} \) = 2ythus critical point is : (0,0)\( \mathrm{{{f}_{{{x}{x}}}}} \) = 2 , \( \mathrm{{{f}_{{{y}{y}}}}} \) = 2 , \( \mathrm{{{f}_{{{x}{y}}}}} \) = 0 , D = (2)(2) - 0 = 4 > 0,hence absolute minimum will be obtained at (0,0)putting it into the function we get ,absolute minimum = f(0,0) = 0 Explanation:as we have obtained D(x,y) > 0 hence obtained critical point will be absolute minimumStep2/3f(x,y) = x^2 + y^2checking the boundary : g(x,y) = 3x + 9y = 8using method of LaGrange's multiplier :\( \begin{align*} \mathrm{{{f}_{{{x}}}}} &= \mathrm{\lambda{{g}_{{{x}}}}} \end{align*} \)2x = 3\( \mathrm{\lambda} \)x = \( \mathrm{\frac{{{3}\lambd ... See the full answer