+0  
 
0
143
3
avatar

1) Suppose the function f(x,y,z)=xyz is defined for x+y+z=7, x,y,z \(\ge \) 0. What is the range of f?

2) The arithmetic mean, geometric mean, and harmonic mean of a, b, c are 8, 5, 3 respectively. What is the value of \(a^2+b^2+c^2\)?

 Aug 30, 2019
 #1
avatar+23878 
+2

2)
The arithmetic mean, geometric mean, and harmonic mean of a, b, c
are 8, 5, 3 respectively.
What is the value of \(a^2+b^2+c^2\)?

 

\(\text{$(1)\ $Arithmetic mean: $\dfrac{a+b+c}{3}=8$ } \\ \text{$(2)\ $Geometric mean: $ \sqrt[3]{abc}=5$ } \\ \text{$(3)\ $Harmonic mean: $ \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} = 3 $ }\)

 

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{a+b+c}{3} &=&8 \\ & a+b+c &=& 3\cdot 8 \\ &\mathbf{a+b+c} &=& \mathbf{24} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2) & \sqrt[3]{abc} &=& 5 \\ & abc &=& 5^3 \\ &\mathbf{abc} &=& \mathbf{125} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& 3 \\\\ & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}&=& \dfrac{3}{3} \\\\ &\mathbf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (4) & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=& 1 \quad | \quad \times abc \\ & \dfrac{abc}{a}+\dfrac{abc}{b}+\dfrac{abc}{c} &=& abc \\ & \mathbf{bc+ac+ab} &=& \mathbf{abc} \\\\ (5) & (a+b+c)^2 &=& a^2+b^2+c^2 + 2(\mathbf{bc+ac+ab}) \\ & (a+b+c)^2 &=& a^2+b^2+c^2 + 2abc \quad | \quad a+b+c = 24,\ abc = 125 \\ & 24^2 &=& a^2+b^2+c^2 + 2\times 125 \\ & 576 &=& a^2+b^2+c^2 + 250\\ & a^2+b^2+c^2 &=& 576-250 \\ & \mathbf{a^2+b^2+c^2} &=& \mathbf{326} \\ \hline \end{array}\)

 

laugh

 Aug 31, 2019
 #2
avatar+106535 
+2

2)

 

Arithmetic mean  =    a + b + c

                                 ________  =  8    →  a + b + c  = 24   →  b + c = 24 - a     (1)

                                       3

 

Geometric mean  =       ∛[abc] = 5     →   abc   = 125

 

Harmonic mean =           3                                                1                                  

                                ____________   =    3        →   __________      =  1    → 

                                 1    +    1    + 1                           bc + ac + ab                      

                                __         __     __                         ____________

                                 a          b        c                                  abc

 

 

      abc

__________  =   1     .....so.....

bc + ac + ab

 

abc  =   bc + ac + ab

125  = a(b + c) + bc

125 - bc  = a (b + c)

125 - bc = a (24 - a)

250 - 2bc  = 2a(24 - a)        (2)

 

Square both sides of (1)  →  b^2 + 2bc + c^2  =  (24 - a)^2    (3)

 

Add (2)  + (3)

 

b^2 + 2bc + c^2   =  (24 - a)^2  

250  - 2bc          =   2a (24 - a)

__________________________________

 b ^2 + c^2 + 250 =  (24 - a)^2  + 2a(24 - a)

 

b^2 + c^2 + 250   =  a^2 - 48a + 576  + 48a  - 2a^2

 

b^2 + c^2 + 250 =  -a^2 + 576

 

a^2 + b^2 + c^2    =  326 

 

 

cool cool cool

 Aug 31, 2019
 #3
avatar+23878 
+2

1)
Suppose the function \(f(x,y,z)=xyz\)
is defined for
\(x+y+z=7, x,y,z\geq 0\).
What is the range of \(f(x,y,z)\)?

 

\(\begin{array}{|rcll|} \hline f(x,y,z) &=& xyz \quad | \quad x+y+z=7 \text{ or } z = 7-x-y \\ &=& xy(7-x-y) \\\\ f(x,y) &=& xy(7-x-y) \\ f(x,y) &=& 7xy-x^2y-xy^2 \\ \hline f_x = \dfrac{\partial f(x,y)}{\partial x} &=& 7y-2xy-y^2 \\ f_y = \dfrac{\partial f(x,y)}{\partial y} &=& 7x-x^2-2xy \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & f_x=0 &=& 7y-2xy-y^2 \\ (2) & f_y=0 &=& 7x-x^2-2xy \\ \hline (1) & y(7-2x-y) &=& 0 \\ & \mathbf{y} &=& \mathbf{0} \\\\ & 7-2x-y &=& 0 \\ & \mathbf{y} &=& \mathbf{7-2x} \\ \hline (2) & x(7-x-2y) &=& 0 \\ & \mathbf{x} &=& \mathbf{0} \\\\ & 7-x-2y &=& 0 \\ & 7-x-2(y) &=& 0 \\ & 7-x-2(0) &=& 0 \\ & 7-x &=& 0 \\ & \mathbf{x} &=& \mathbf{7} \quad | \quad y=0 \\\\ & 7-x-2(y) &=& 0 \\ & 7-x-2(7-2x) &=& 0 \\ & 7-x-14+4x &=& 0 \\ & 3x &=& 7 \\ & \mathbf{x} &=& \mathbf{\dfrac{7}{3} } \quad | \quad y=7-2x \\\\ \hline & y &=& 7-2x \\ & y &=& 7-2(0) \\ & \mathbf{y} &=& \mathbf{7} \quad | \quad x=0 \\\\ & y &=& 7-2x \\ & y &=& 7-2(7) \\ & \mathbf{y} &=& \mathbf{-7} \quad | \quad x=7 \\\\ & y &=& 7-2x \\ & y &=& 7-2(\dfrac{7}{3}) \\ & \mathbf{y} &=& \mathbf{\dfrac{7}{3} }\quad | \quad x=\dfrac{7}{3} \\ \hline \end{array}\)

 

\(\begin{array}{|c|c|c|c|c|r|} \hline x&y&z=7-x-y & \text{solution} & f(x,y,z) & \\ \hline 0 &7 &0&&0&\text{minimum} \\ 7 &0 &0&&0&\text{minimum} \\ 7 &-7 &0& \text{no }(y\geq 0!)&& \\ \dfrac{7}{3} & \dfrac{7}{3} & \dfrac{7}{3} && \dfrac{343}{27} & \text{maximum} \\ \hline \end{array}\)

 

Range of \(f(x,y,z):\ 0\ldots \dfrac{343}{27}\)

 

laugh

 Sep 1, 2019

19 Online Users

avatar