Question Suppose that \( f(x, y)=x^{2}+y^{2} \) at which \( 0 \leq x, y \) and \( 3 x+9 y \leq 8 \). Please enter exact answers. 1. Absolute minimum of \( f(x, y) \) is 2. Absolute maximum of \( f(x, y) \) is
【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