Question

The monthly profit of Sweet Chew Chocolate Ltd. is represented as a function of monthly sales S in units of the small size bar x, and the big size bar y sold each month. Specifically, the function is of the form: S = a x² + by? + cxy (where a, b and c are constants). Observations in recent past indicate that monthly sales were: €50,950 when 500 and 1,000 units of small and large bars respectively were sold; £377,500 when 1500 and 2,500 units respectively were sold; and £603, 300 when 2,000 and 3,000 units respectively were sold. Determine: (a) Sales function (b) Sales when 1,000 and 2,000 units of small and large bars respectively are sold.

Answer 1

Answer

a)

$S = 0.03x^2 + 0.0127y^2 + 0.0615xy$

b) $S = £203,800$

Workings

Since a general Sales function is provided, the best approach to the problem is to substitute the given values of x and y into the general function to produce a system of equations. The equation system is then solved simultaneously to obtain the values of constants a, b and c.

It is given that when x = 500, and y = 1,000, Sales = £50,950.

Therefore,

£50,950 = $a(500)^2 + b(1,000)^2 + c(500)(1,000)$

£50,950 = 250,000a + 1,000,000b + 500,000c

Dividing the equation by 1,000 to reduce it will give:

$£50.95 = 250a + 1,000b + 500c --- equation 1$

Also, when x = 1,500; b = 2,500 and Sales = £377,500

Thus,

£377,500 = $a(1,500)^2 + b(2,500)^2 - c(1,500)(2,500)$

£377,500 = 2,250,000a + 6,250,000b + 3,750,000c

Dividing the equation by 1,000 to reduce it gives:

$£377.50 = 2,250a + 6,250b + 3,750c ---equation 2$

Also, Sales = £603,300 when x = 2,000 and y = 3,000

So,

£603,300 = $a(2,000)^2 + b(3,000)^2 + c(2,000)(3,000)$

£603,300 = 4,000,000a + 9,000,000b + 6,000,000c

Reducing the equation by dividing it by 1,000 produces:

$£603.30 = 4,000a + 9,000b + 6,000c --- equation 3$

Solving the system of equations

Multiplying equation 1 by 9 gives:

$£458.55 = 2,250a + 9,000b + 4,500c --- equation 4$

Subtracting equation 2 from equation 4

£458.55 = 2,250a + 9,000b + 4,500c

-(£377.50 = 2,250a + 6,250b + 3,750c)

=> $2,750b + 750c = £81.05 --- equation 5$

Multiplying equation 1 by 16 gives:

$4,000a + 16,000b + 8,000c = £815.20 --- equation 6$

Subtracting equation 3 from equation 6:

4,000a + 16,000b + 8,000c = £815.20

-(4,000a + 9,000b + 6,000c = £603.30)

=> $7,000b + 2,000c = £211.90 --- equation 7$

Utilizing equations 7 and 5:

$7,333.333333b + 2,000c = £216.13 --- equation 5 \space × \dfrac {8} {3}$

$-(7,000b + 2,000c = £211.90) --- equation 7 \space × 1$

=> $333.333333b = 4.233333$

=> $b = \dfrac {4.233333333} {333.333333}$

=> $b = 0.0127$

Substituting 0.0127 for b in equation 7:

7,000(0.0127) + 2,000c = 211.90

=> $c = \dfrac {211.90 - 88.90} {2,000}$

=> $c = \dfrac {123}{2,000}$

=> $c = 0.0615$

Substituting values of b and c in equation 1:

250a + 1,000(0.0127) +500(0.0615) = £50.95

$=> a = \dfrac {50.95 - 12.7 - 30.75} {250}$

$=> a = \dfrac {7.50} {250}$

$=> a = 0.03$

Substituting values of a, b and of c into the general function gives:

$$S = 0.03x^2 + 0.0127y^2 + 0.0615xy$$

b) When x = 1,000 and y = 2,000

$S = 0.03(1,000)^2 + 0.0127(2,000)^2 +0.0615(1,000 × 2,000)$

$S = 30,000 + 50,800 + 123,000$

$S = £203,800$