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

$$\begin {align*} x + y &\geq 10,\\
2x + 3y &\geq 20,\\
x &\geq 4,\\
y &\geq 4.
\end{align*}$$

  1. $32$
  2. $50$
  3. $52$
  4. $100$
  5. None of the above
asked in Calculus by Boss (29.5k points) | 406 views

4 Answers

+6 votes
Best answer
Answer: Option $(B)$ $50$

 $x\geq 4$ and $y\geq 4$ , So we can take both $x =5$ & $y = 5$

$x+y \geq 10,$ Satisfied , $5+5 = 10$

$2x + 3y \geq 20$. Satisfied.

This is in fact the minimum value.

Other options:

$4,4 \implies x+y \geq 10$ constraint fail

$4,5 \implies  x+y \geq 10$ fail

$6,4 \implies$ Sum $=52$ which is more than $50$ and hence cannot  be answer.

$7,3 \implies 49+9 = 58$ which is again greater than $50$ and hence cannot be the answer.
answered by Boss (41k points)
edited by
0
is brute force only possible way to solve such questions????
+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
answered by Boss (15.8k points)
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
answered by Active (2.6k points)
0
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
–1 vote

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

answered by Active (3.7k points)
+1

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

0

@Arjun ,

why 52 is answer ? It should be 50.

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