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Zorluk: Çok zorOrder of Operations and Number Properties

If xx, yy, and zz are integers such that −5≤x<y<z≤5-5 \leq x < y < z \leq 5, what is the minimum possible value of the expression (x−y)2−z(y−x)−∣x∣⋅∣y−z∣(x - y)^2 - z(y - x) - |x| \cdot |y - z|?

  1. A
    -1
  2. B
    -25
  3. C
    -39
  4. D
    -41
  5. -49Cevap

Cevap

The minimum possible value of the expression is −49-49.
The correct value is −49-49. By setting d1=y−x≥1d_1 = y - x \geq 1 and d2=z−y≥1d_2 = z - y \geq 1, we can rewrite the expression as d12−zd1−∣x∣d2d_1^2 - z d_1 - |x| d_2. Substituting z=x+d1+d2z = x + d_1 + d_2 gives −xd1−d2(d1+∣x∣)-x d_1 - d_2(d_1 + |x|). To minimize this, we choose the smallest possible value for xx, which is −5-5. This gives 5d1−d2(d1+5)5d_1 - d_2(d_1 + 5). Since x=−5x = -5 and z≤5z \leq 5, we have d1+d2≤10d_1 + d_2 \leq 10. Substituting d2=10−d1d_2 = 10 - d_1 yields 5d1−(10−d1)(d1+5)=d12−505d_1 - (10 - d_1)(d_1 + 5) = d_1^2 - 50. The minimum value of this quadratic expression for integer d1≥1d_1 \geq 1 occurs at d1=1d_1 = 1, giving 12−50=−491^2 - 50 = -49. This corresponds to x=−5,y=−4,z=5x = -5, y = -4, z = 5. Evaluating the expression directly confirms this result: (−5−(−4))2−5(−4−(−5))−∣−5∣⋅∣−4−5∣=1−5−45=−49(-5 - (-4))^2 - 5(-4 - (-5)) - |-5| \cdot |-4 - 5| = 1 - 5 - 45 = -49.

Adım Adım Çözüm

1
Define variables for the differences between the ordered integers.
Let d1=y−x≥1d_1 = y - x \geq 1 and d2=z−y≥1d_2 = z - y \geq 1, which implies y=x+d1y = x + d_1 and z=x+d1+d2z = x + d_1 + d_2.
Using differences simplifies the inequality constraints and allows us to express the objective function in terms of positive integers d1d_1 and d2d_2.
2
Substitute the differences into the given expression (x−y)2−z(y−x)−∣x∣⋅∣y−z∣(x - y)^2 - z(y - x) - |x| \cdot |y - z|.
The expression becomes d12−(x+d1+d2)d1−∣x∣d2=−xd1−d2(d1+∣x∣)d_1^2 - (x + d_1 + d_2)d_1 - |x|d_2 = -x d_1 - d_2(d_1 + |x|).
This rewrites the expression in terms of the initial variable xx and the positive differences d1d_1 and d2d_2.
3
Analyze how to minimize −xd1−d2(d1+∣x∣)-x d_1 - d_2(d_1 + |x|) given the boundaries.
Setting x=−5x = -5 makes the expression 5d1−d2(d1+5)5d_1 - d_2(d_1 + 5). Since x=−5x = -5 and z≤5z \leq 5, we have d1+d2≤10d_1 + d_2 \leq 10.
Choosing the minimum value for xx maximizes the positive coefficient of d1d_1 and the term ∣x∣|x| in the negative product, leading to the smallest possible value.
4
Maximize d2d_2 by setting d2=10−d1d_2 = 10 - d_1 and substitute it into the expression.
The expression becomes 5d1−(10−d1)(d1+5)=d12−505d_1 - (10 - d_1)(d_1 + 5) = d_1^2 - 50.
Since d2d_2 has a negative coefficient, maximizing d2d_2 minimizes the overall expression.
5
Minimize d12−50d_1^2 - 50 subject to d1≥1d_1 \geq 1.
The minimum occurs at d1=1d_1 = 1, yielding 12−50=−491^2 - 50 = -49, which corresponds to x=−5,y=−4,z=5x = -5, y = -4, z = 5.
Since d12d_1^2 is strictly increasing for positive integers, the minimum value of the quadratic is achieved at the smallest boundary value of d1d_1.

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