Question 1086875
.
2x^2 + y^2 = 4
the max and min value of 4x + y^2 ?
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You are given that  

{{{2x^2 + y^2}}} = {{{4}}},    (1)

and they ask you to find max and min of

{{{4x + y^2}}}.          (2)


    Notice that, due to (1), the domain for x is the set of real numbers  {{{-sqrt(2)}}} <= x <= {{{sqrt(2)}}}.


From (1), you have {{{y^2}}} = {{{4 - 2x^2}}}.   Substitute it into (2), replacing  {{{y^2}}}.  You will get

{{{4x + y^2}}} = {{{4x + 4 - 2x^2}}} = {{{-2*(x^2 - 2x) + 4)}}} = (completing the square) = {{{-2*(x^2 - 2x + 1) + 2 + 4)}}} = {{{-2*(x-1)^2 + 6}}}.


Thus the maximum of  {{{4x + y^2}}}  is equal to  6  and it is achieved at x = 1.


Or, more precisely, the maximum of  {{{4x + y^2}}}  is achieved at the points  (x,y) = ({{{1}}},{{{sqrt(2)}}})  and  (x,y) = ({{{1}}},{{{-sqrt(2)}}}).


The plot below shows the graph  f(x) = {{{-2*(x-1)^2 + 6}}} in the interval  [{{{-sqrt(2)}}},{{{sqrt(2)}}}].


{{{graph( 330, 330, -1.5, 1.5, -10.5, 10.5,
          -2*(x-1)^2 + 6
)}}}


Plot  f(x) = {{{-2*(x-1)^2 + 6}}}



The minimum is at x = {{{-sqrt(2)}}} and is equal to {{{-2*(-sqrt(2)-1)^2 + 6}}} = {{{-2*(1+sqrt(2))^2 + 6}}} = -5.65 (approximately).
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