<p>Test 4 Pg.66 #25....</p>
<p>If x is an interger between the interval -2<x<4 and y is an interger between the interval -3<equalto y<0, then what is the maxium value of (x+y)^2=? </p>
<p>A)0
B)1
C)4
D)9
E)16</p>
<p>Book Answer: C ...***?</p>
<p>Why can't it be X max=-1 Y max = -3 ((-3)+(-1))^2 =-4^2=16 ?</p>
<p>prolly just a mistake</p>
<p>Hm, thats good to know, because I thought I was doing something wrong.</p>
<p>it doesnt seem wrong because it makes sense, unless u copied the wrong ranges down. Otherwise all the answer choices are possible and 16 does work</p>
<p>yes, lots of mistakes in his book. Ironic, really, since he comes up with creative and challenging problems, yet makes basic mistakes.</p>
<p>Note: Rusen Meylani is on the forums here, saw him in another thread. his username is meylani.</p>
<p>Dear Carbon,</p>
<p>That is a mistake that we are avare of.</p>
<p>Best of luck in the exam.</p>
<p>Rusen Meylani.</p>
<p>could someone work this out to show what ?<(x+y)^2<? is</p>
<p>i always get confused on these for some reason</p>
<p>Slipstream99:</p>
<p>You need to choose values for x and y so that ABS(x+y) is as large as possible. One way is to try all 4 combinations of</p>
<p>min(x) + min(y)
min(x) + max(y)
max(x) + min(y)
max(x) + max(y)</p>
<p>and see which one of these gives the highest value for ABS(x+y). For the problem shown, you would get</p>
<p>-2-3, -2+0, 4-3, 4+0
or -5, -2, +1, +4</p>
<p>The first one has the max ABS() value; choosing x=-2 (or -1.9999999, if you want to be picky) and y=-3 would give
(x+y)^2 = (-2 -3)^2 = 25, which is none of the answers shown. </p>
<p>Looks like the answers need to be reviewed...</p>
<p>thanks, that clears up why i was getting 25 and thought I was wrong...
but how do you do it without all that checking. </p>
<p>as in...
?<x+y<? =
-2 + -3 = -5<x+y<4 = 0 + 4</p>
<p>i just don't know how to do the squaring, because isn't the minimum value 0?</p>
<p>also, do you have to turn around inequality signs in squaring or anything besides multiplying and dividing negative numbers?</p>
<p>Slipstream99:
Careful, now - the minimum value of any ()^2 is zero, assuming we ignore imaginary numbers. However, if you have an inequality like</p>
<p>-5 < x + y</p>
<p>then x+y can take on values like -4.95, -3.2, -1, -0.3, 0, +1, +4578,....</p>
<p>and (x+y)^2 would take on approx. values of 24.5, 9.6, 1, 0.09, 0, 1, ... and the smallest value possible for it would be the 0.</p>
<p>x and y both are INTEGERS.</p>
<p>The right answer is E. The value (x+y)^2 is maximum when x=-1, y=-3.</p>
<p>that explains it, thanks!</p>