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Solve min $x^{2}+y^{2}$

subject to

$x + y \geq 10,$

$2x + 3y \geq 20,$

$x \geq 4,$

$y \geq 4.$

1. $32$
2. $50$
3. $52$
4. $100$
5. None of the above
asked in Calculus | 205 views

x>=4 and y>=4 , So we can take both x =5 & y = 5

x+y >= 10 => Satisfied , 5+5 = 10

2x + 3y >= 20. Satisfied.

This is infact minimum value.

Other options =>

4,4 => x+y constraint fail

4,5 => x+y fail

6,4 => Still giving 52 as sum which is more than 50 !,  This can not  be answer.

7,3 => 49+9 > 58 > 50.
selected
+1 vote
I think it's the easiest one,
approach - first we just break to the minimum conditions so every thing can be meet. minimum value such that every thing can satisfy is 5 , 5 which gives = 50,

now all the possiblities are decreasing one number and increasing one. but as u think . as u will decrease one number the value that will decrease will be less then the after effect of increasing the other number.
like 4 and 6 . value that decreases due to decreasing it from 5 is (25-16) = 9

but the increase in the value due to increasing 5 to 6 is (36-25)= 11 , so the best answer will be the mid point i.e 5,5 = 50

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

why B is not the ans

because minimum of min(x+ 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(x+ y2) = 25+25 = 50

why 52 is answer ? It should be 50.

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