Winter 2015 – Q.3

Q.3 a) Determine the value of a and b such that system $latex \left\{\begin{array}{l}\begin{array}{l}3x-2y+z=b\\5x-8y+9z=3\end{array}\\\;\;2x+y+az=-1\end{array}\right.$

has i) no solution, ii) a unique solution, iii) infinite number of solutions [6M]

$latex \begin{array}{l}3x-2y+z=b\\\\5x-8y+9z=3\\\\2x+y+az=-1\\\\\\\\\\\\\\\\\\\end{array}$

$latex \begin{array}{l}\begin{bmatrix}3&-2&1\\5&-8&9\\2&1&a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}b\\3\\-1\end{bmatrix}=Ax=B\\\\\\\\\\\\\\\\\\\end{array}$

$latex \begin{array}{l}c\sim\begin{pmatrix}3&-2&1&.&b\\5&-8&9&.&3\\2&1&a&.&1\end{pmatrix}R_1/3=R_1\\\\\\\\\\\\\\\\\\\end{array}$

$latex \sim\begin{bmatrix}1&-2/3&1/3&.&b/3\\5&-8&9&.&3\\2&1&a&.&1\end{bmatrix}
$latex R_2=R_2-5R,R_3=R_3-2R$

$latex \sim\begin{bmatrix}1&-2/3&y3&.&b/3\\0&-14/3&22/3&.&3-\frac{5b}3\\0&7/3&\frac{3a-2}3&.&1-\frac{2b}3\end{bmatrix}R_3=R_3+\frac{R_2}2$

$latex \sim\begin{bmatrix}1&-2/3&y3&.&b/3\\0&-14/3&22/3&.&\frac{9-5b}3\\0&0&a+3&.&\frac{5-3b}2\end{bmatrix}$

1) for no (error)

$latex if\;R\left(A\right)\neq R\left(C\right)$

$latex \begin{array}{l}i.e.a+3=0\;,\;a=-3\;\&\\\\\\\frac{5-3b}2\neq0\;\;,5\neq3b\\\\b\neq5/3\end{array}$

2)for unique soln.

$latex \begin{array}{l}R\left(A\right)=R\left(C\right)=3\\\\i.e.a+3\neq0\;,a\neq-3\\\\\end{array}$

and b may have any value

3)for unique soln

$latex \begin{array}{l}R\left(A\right)=R\left(C\right)=2\\\\a+3=0,a=-3\;and\\\\\frac{5-3b}2=0,5=3b\;,b=5/3\\\\\\\\\end{array}$

b) Discuss the maximum and minimum of $latex \left(x,\;y\right)=x^3+3xy^2-15\left(x^2+y^2\right)+72x$ [6M]

$latex \begin{array}{l}\frac{\partial f}{\partial x}=3x^2+3y^2-30x+72\\\\\\\\\\\end{array}$

$latex \begin{array}{l}\frac{\partial f}{\partial y}=6xy-30y\\\\\\\\\\\end{array}$

$latex \begin{array}{l}\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0\\\\\\\\\\\end{array}$

$latex \begin{array}{l}3x^2+3y^2-30x+72=0……1\\\\6xy-30y=0\\\\6y\left(x-5\right)=0\\\\\\\\\\\end{array}$

x=5
y=0

put x=5 in 1, we get

$latex \begin{array}{l}3\times25+3y^2-150+72=0\\\\3y^2=3\\\\y\neq1\\\\\end{array}$

put y=0 in 1

$latex \begin{array}{l}3x^2-30x+72=0\\\\x^2-10x+24=0\\\\\left(x-6\right)\left(x-y\right)=0\\\\\\\\\end{array}$

x=6
x=4

when x=5,y=(error)1

y=0,x=6,4

$latex \begin{array}{l}r=\frac{\partial^2f}{\partial x^2}=6x-30\\\\t=\frac{\partial^2f}{\partial y^2}=6x-30\\\\s=\frac{\partial^2f}{\partial x\partial y}=6y\\\\\\\\\end{array}$

when $latex \begin{array}{l}x=5,y=\pm1\\\\\\\\\\\end{array}$

$latex \begin{array}{l}rt-s^2=\left(6\times5-30\right)\left(6\times5-30\right)-36\times\left(\pm1\right)^2\\\;\;\;\;\;\;\;\;\;\;\\\;\;\;\;\;\;\;\;\;\;=-36<0=-ve\\\\\\\\\\\end{array}$ when x=6,y=0 $latex \begin{array}{l}rt-s^2=\left(36-30\right)\left(36-30\right)\\\\\;\;\;\;\;\;\;\;\;\;=36>0=+ve\\\\\\\\\\\end{array}$

$latex \begin{array}{l}r=6x-30,x=6\\\\\;\;=6>0=minimum\;value\\\\\\\\\\\end{array}$

$latex \begin{array}{l}r=6x-30,x=4\\\;\;=24-30=-6<0=maximum\;value\\\\\\\\\\\end{array}$ Hnce,f(x,y) is minimum a (4,0) and maximum a (4,0)

c) show that $latex \tan^{-1}\left(\frac{x-ty}{x-ty}\right)=\frac\pi4+\frac t2\log\left(\frac{x+y}{x-y}\right)$ [8M]