9 votes 9 votes Solve min $x^{2}+y^{2}$ subject to $$\begin {align*} x + y &\geq 10,\\ 2x + 3y &\geq 20,\\ x &\geq 4,\\ y &\geq 4. \end{align*}$$ $32$ $50$ $52$ $100$ None of the above Calculus tifr2014 calculus maxima-minima + – makhdoom ghaya asked Nov 9, 2015 edited Aug 15, 2020 by soujanyareddy13 makhdoom ghaya 1.7k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes x+y≥10 2x+3y≥20, x≥4, y≥4. we have to choose that value of x & y which satifies all the equations, as well as give the min ( x^ 2 + y^ 2) which can be possible when X= 5 and Y=5 , min ( x^ 2 + y^ 2) = 5^2 + 5^2 = 50 priyanka gautam-piya answered Dec 14, 2016 priyanka gautam-piya comment Share Follow See all 2 Comments See all 2 2 Comments reply Kaluti commented Sep 14, 2017 reply Follow Share but (5,5) point does not lie in bounded region of this lpp how we are saying (5,5) will be point of convex region should not the answer be (4,6) which gives 52 as answer 0 votes 0 votes ankitgupta.1729 commented Oct 2, 2020 reply Follow Share @Kaluti, it is not lpp. Here, objective function is not linear. 0 votes 0 votes Please log in or register to add a comment.
–1 votes –1 votes option c) 52 Here , we have constraints x>=4 and y>=4 In order to satisfy the equation x+y >= 10 , we need to have min value of x and y as 4 and 6 . so , min(x2 + y2) = 16+36 = 52 worst_engineer answered Nov 11, 2015 worst_engineer comment Share Follow See all 2 Comments See all 2 2 Comments reply tiger commented Nov 28, 2015 reply Follow Share why B is not the ans because minimum of min(x2 + y2) Here , we have constraints x>=4 and y>=4 In order to satisfy the equation x+y >= 10 , we need to have min value of x and y as 5 and 5. so , min(x2 + y2) = 25+25 = 50 1 votes 1 votes Akash Kanase commented Dec 6, 2015 reply Follow Share @Arjun , why 52 is answer ? It should be 50. 0 votes 0 votes Please log in or register to add a comment.