What is the maximum value of x^2+2y^2+5x-35? It is known that x^2+y^2=5.
A.22 B.20 C.0 D.-20 E.-22
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58 find the max value of expression
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cbenk121
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Tricky problem.
So here is my system of equations:
x^2 + 2y^2 + 5x - 35 = 0
x^2 + y^2 = 5
Neither X nor Y can be much larger than 2, because it would exceed 5.
So I plugged in X = 3; Y = 0 into the top equation
3^2 + 2*0^2 + 5(3) = 35 (It would be 35 if max value was 0)
24 = 35
OK, so now I know that it can not be (A), (B), or (C). This is because when X is outside of constraint, the value still can not reach zero.
If we try X=2, and Y=1, then we get -19. If we try X=1 and Y=2, then we get -21.
So at this point, I'm not sure how to proceed. Assuming the information given was correct, I'd probably guess (D) since I found two values above -22 and move on.
So here is my system of equations:
x^2 + 2y^2 + 5x - 35 = 0
x^2 + y^2 = 5
Neither X nor Y can be much larger than 2, because it would exceed 5.
So I plugged in X = 3; Y = 0 into the top equation
3^2 + 2*0^2 + 5(3) = 35 (It would be 35 if max value was 0)
24 = 35
OK, so now I know that it can not be (A), (B), or (C). This is because when X is outside of constraint, the value still can not reach zero.
If we try X=2, and Y=1, then we get -19. If we try X=1 and Y=2, then we get -21.
So at this point, I'm not sure how to proceed. Assuming the information given was correct, I'd probably guess (D) since I found two values above -22 and move on.
