[amp_mcq option1=”is a minimum equal to \\[\\frac{{10}}{3}\\]” option2=”is a maximum equal to \\[\\frac{{10}}{3}\\]” option3=”is a minimum equal to \\[\\frac{8}{3}\\]” option4=”is a maximum equal to \\[\\frac{8}{3}\\]” correct=”option1″]<\/p>\n
The correct answer is: A. is a minimum equal to $\\frac{{10}}{3}$.<\/p>\n
To find the optimal value of a function, we can use the method of Lagrange multipliers. The Lagrange function for this problem is:<\/p>\n
$$L(x, y, \\lambda) = 4x^2 + 6y^2 – 8x – 4y + 8 + \\lambda(4x – 8y)$$<\/p>\n
Setting the partial derivatives of $L$ with respect to $x$, $y$, and $\\lambda$ equal to zero, we get the system of equations:<\/p>\n
$$\\begin{align} Solving this system of equations, we get the critical points $(x, y) = \\left(\\frac{1}{2}, \\frac{1}{3}\\right)$.<\/p>\n To evaluate the function $f$ at the critical point, we get:<\/p>\n $$f\\left(\\frac{1}{2}, \\frac{1}{3}\\right) = \\frac{{10}}{3}$$<\/p>\n Since the critical point is a minimum, we know that the optimal value of $f$ is $\\frac{{10}}{3}$.<\/p>\n Here is a brief explanation of each option:<\/p>\n [amp_mcq option1=”is a minimum equal to \\[\\frac{{10}}{3}\\]” option2=”is a maximum equal to \\[\\frac{{10}}{3}\\]” option3=”is a minimum equal to \\[\\frac{8}{3}\\]” option4=”is a maximum equal to \\[\\frac{8}{3}\\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20176","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n
\n8x – 8\\lambda &= 0 \\
\n6y – 8\\lambda &= 0 \\
\n4x – 8y &= 0
\n\\end{align<\/em>}$$<\/p>\n\n