Question

A motor company manufacture and sell cars and motorbikes. The cost of manufacturing x motorbikes and y cars is given by 2 2 C x y x xy y ( , ) 100 100 400 = + + . Each motorbike is sold for N$36 000-00 and each car is sold for N$180 000-00. Use Cramer’s rule to determine the number of motorbikes and the number of cars that should be manufactured and sold for a maximum profit  and determine the maximum profit max . (10 marks) 4.2 Use the Jacobian to test for functional dependence between the cost and the revenue functions in 4.1. (7 marks) 4.3 One of the stationary points of the function ( ) 4 4 2 2 f x y x y x xy y , 2 4 2 = + − + − is ( 2, 2 − ). Use the Hessian to test whether the given point is a maximum, minimum or a saddle point.

Answer 1

4.1

Cost of manufacturing ,C(x,y)=$100x^2+100xy+400y^2$ where x is the number of motorbikes and y the number of cars.

$\therefore$ Revenue,R(x,y)=$36,000x+180,000y$

$\therefore$ profit, $\pi$ =$36,000x+180,000y-100x^2-100xy-400y^2$

For $\pi$ max,$\frac{\delta\pi}{\delta x}=36,000-200x-100y=0$

$2x+y=360$ .........................................(1)

$\frac{\delta\pi}{\delta y}=180,000-100x-800y=0$

$x+8y=1800$ ........................................(2)

Using Gaussian elimination to eliminate x;

$(2x+16y)-(2x+y)=3600-360$

$15y=3240$

$y=216$

Putting this value of y in (2)

$x=1800-1728=72$

$\therefore$ for maximum profit, $\pi$ ,no of motorbikes=72 and no of cars=216.

Maximum profit =36,000(72)+180,000(216)-100(72)$^{2}$ -100(72)(216)-400(216)$^{2}$ =$\$20,736,000$

4.2

$\pi(x,y)=36,000x+180,000y-100x^2-100xy-400y^2$

$\frac{\delta^{2}\pi}{\delta x^{2}}=\frac{\delta}{\delta x}36,000-200x-100y=-200<0$

$\frac{\delta^{2}\pi}{\delta y^{2}}=-800$

$\frac{\delta^{2}\pi}{\delta x\delta y}=-100$

$\frac{\delta^{2}\pi}{\delta y\delta x}=-100$

Hessian, H $\begin{vmatrix} \frac{\delta^{2}\pi}{\delta x^{2}}& \frac{\delta^{2}\pi}{\delta y\delta x} \\ \frac{\delta^{2}\pi}{\delta x\delta y}& \frac{\delta^{2}\pi}{\delta y^{2}} \end{vmatrix}$ =$\begin{vmatrix} -200 & -100 \\ -100 & -800 \end{vmatrix}$

The above solution offers maximum profit.

4.3

To test for functional dependence between the cost and the revenue functions in 4.1 the Jacobian

= $\frac{\delta\ (C, R)}{\delta (x, y)}$ should be zero.

$\frac{\delta\ (C, R)}{\delta (x, y)}=$ $\begin{vmatrix} \frac{\delta C}{\delta x}& \frac{\delta C}{\delta y} \\ \frac{\delta R}{\delta x}& \frac{\delta R}{\delta y} \end{vmatrix}$ =$\begin{vmatrix} 200x+100y & 100x+800y \\ 36,000& 180,000 \end{vmatrix}$

=36000000$x$ [phone]$y$ -3600000$x$ -28800000$y$

=32400000$x$ -10800000$y$

=10800000$(3x-y)$

$\frac{\delta\ (C, R)}{\delta (x, y)}$ =10800000$(3\times72-216)=0$

$\therefore$ cost and revenue functions are functionality dependent.