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 \(a^2+b^2+c^2\)?
+26367 2)
The arithmetic mean, geometric mean, and harmonic mean of a, b, c are 8, 5, 3 respectively.
What is the value of \(a^2+b^2+c^2\)?
\(\text{$(1)\ $Arithmetic mean: $\dfrac{a+b+c}{3}=8$ } \\ \text{$(2)\ $Geometric mean: $ \sqrt[3]{abc}=5$ } \\ \text{$(3)\ $Harmonic mean: $ \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} = 3 $ }\)
\(\begin{array}{|lrcll|} \hline (1) & \dfrac{a+b+c}{3} &=&8 \\ & a+b+c &=& 3\cdot 8 \\ &\mathbf{a+b+c} &=& \mathbf{24} \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline (2) & \sqrt[3]{abc} &=& 5 \\ & abc &=& 5^3 \\ &\mathbf{abc} &=& \mathbf{125} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (3) & \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& 3 \\\\ & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}&=& \dfrac{3}{3} \\\\ &\mathbf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}} &=& \mathbf{1} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (4) & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=& 1 \quad | \quad \times abc \\ & \dfrac{abc}{a}+\dfrac{abc}{b}+\dfrac{abc}{c} &=& abc \\ & \mathbf{bc+ac+ab} &=& \mathbf{abc} \\\\ (5) & (a+b+c)^2 &=& a^2+b^2+c^2 + 2(\mathbf{bc+ac+ab}) \\ & (a+b+c)^2 &=& a^2+b^2+c^2 + 2abc \quad | \quad a+b+c = 24,\ abc = 125 \\ & 24^2 &=& a^2+b^2+c^2 + 2\times 125 \\ & 576 &=& a^2+b^2+c^2 + 250\\ & a^2+b^2+c^2 &=& 576-250 \\ & \mathbf{a^2+b^2+c^2} &=& \mathbf{326} \\ \hline \end{array}\)
+128408 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
+26367 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 \(f(x,y,z):\ 0\ldots \dfrac{343}{27}\)
