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Calculus Level 3

Given the function f ( r ) = { 0 r = 0 1 r ≠ 0 f(r) = \left\{ \begin{array}{ll} 0 & \quad r= 0 \\ 1 & \quad r \neq 0 \end{array} \right. We define the function g ( x , y , z ) = ( x − 1 ) 2 + ( y − 2 ) 2 + ( z − 9 ) 2 + t [ f ( y − x ) + f ( z − y ) ] , g(x, y, z) = (x-1)^2+(y-2)^2+(z-9)^2+ t [f(y-x)+f(z-y)], where t t represents an arbitrary real number. Assume that the finite interval ( a , b ) (a, b) is the largest open interval that contains all values t t that satisfy the condition that the function g ( x , y , z ) g(x, y, z) has its absolute minimum value at a unique point ( x 0 , y 0 , z 0 ) , (x_0, y_0, z_0), where neither x 0 , x_0, y 0 , y_0, and z 0 z_0 are all equal to each other, nor they are all different from each other. If that interval exists, enter b − a , b-a, otherwise enter 555.


The answer is 37.

Arturo Presa by Arturo Presa , 62, USA

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1 solution

Arturo Presa
May 14, 2020

We are going to consider the following cases.

Case 1 : x = y = z x=y=z

In this case, the minimum of the function g ( x , x , x ) g(x, x, x) is 38 and is attained at the point ( 4 , 4 , 4 ) . (4, 4, 4).

Case 2: x = y x=y and z ≠ x z\neq x

In this case, the minimum value of g ( x , x , z ) g(x, x, z) is t + 0.5 t+0.5 at the point ( 3 / 2 , 3 / 2 , 9 ) . (3/2, 3/2, 9).

Case 3: y = z y=z and x ≠ y x\neq y

In this case, the minimum value of g ( x , y , z ) g(x, y, z) is t + 24.5 t+24.5 at the point ( 1 , 11 / 2 , 11 / 2 ) . (1, 11/2, 11/2).

Case 4: x = z x=z and x ≠ y x\neq y

In this case, the minimum value of g ( x , y , x ) g(x,y, x) is 2 t + 32 2t+32 at the point ( 5 , 2 , 5 ) . (5, 2, 5).

Case 5: x ≠ y , x\neq y, y ≠ z , y\neq z, and x ≠ z x\neq z

In this case, the minimum value of g ( x , y , z ) g(x, y, z) is 2t at the point ( 1 , 2 , 9 ) (1, 2, 9)

According to the problem, the minimum value of the function must happen in Case 2, 3, or 4 only. Since t + 0.5 t+0.5 is always less than t + 24.5 t+24.5 and 2 t + 32 , 2t+32, then it will occur only when the case 2 happens. For Case 1 and Case 5 not to happen, then t + . 5 < 38 t+.5< 38 and t + . 5 < 2 t . t+.5<2t. Therefore, 0.5 < t < 37.5. 0.5<t< 37.5. So the interval is ( 0.5 , 37.5 ) (0.5, 37. 5) and the answer to the problem is 37 . \boxed{37}.

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Sir, but how you found the minimum values at the respective points ? What is the principle of minima and maxima in 3D curves ? I don't know any stuff about it and I am bit confused about the calculus in 3D geometry. Could you please help me about it ?

Pradeep Tripathi - 1 year ago

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The case 1 is very simple because the function g ( x , x , x ) g(x, x, x) is a function of one variable. Then you find the value of x x where the first derivative is zero which is 4. It is a minimum because the second derivative of g ( x , x , x ) g(x, x, x) is positive at x = 4. x=4. The case 5 is also simple because in this case the three variables are different and then you can see that the g ( x , y , z ) = ( x − 1 ) 2 + ( y − 2 ) 2 + ( z − 9 ) 2 + 2 t g(x, y, z)= (x-1)^2+(y-2)^2+(z-9)^2+2t and it is easy to see by inspection that the minimum value of that expression is attained at the point ( 1 , 2 , 9 ) (1, 2, 9) . So, the minimum would be 2t.

For cases 2, 3, and 4, you obtain a function of two variables formed by the sum of two terms, so that each of them depends on only one variable, and third term that depends on another variable.. For example, when x = y x=y and x ≠ z . x\neq z. The function becomes g ( x , x , z ) = ( x − 1 ) 2 + ( x − 2 ) 2 + ( z − 9 ) 2 + t , g(x, x, z)= (x-1)^2+(x-2)^2 +(z-9)^2+ t, where t t is a constant. The minimum of the sum of the first two terms, ( x − 1 ) 2 + ( x − 2 ) 2 , (x-1)^2+(x-2)^2, is attained at x 0 = 3 / 2 x_0=3/2 and the minimum of the other term, ( z − 9 ) 2 (z-9)^2 is attained at z 0 = 9. z_0=9. You can see that x 0 ≠ z 0 . x_0 \neq z_0. Therefore the minimum of g ( x , y , z ) g(x, y, z) is g ( 3 / 2 , 3 / 2 , 9 ) = t + 0.5 g(3/2, 3/2, 9)= t+0.5 in this particular case.

So, you don't have to know optimization of functions of several variables to understand these calculations.

Arturo Presa - 1 year ago

Correct :-) Nice problem!

István Blahota - 1 year ago

Thank you, Istvan!

Arturo Presa - 1 year ago

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