Question

ii. A multinational refreshments firm has 68 monetary units available to produce the maximum possible number of bottles. Its production function is q(x, y) = 60x + 90y − 2x 2 − 3y 2 where x and y are the required inputs. The inputs prices are px = 2 m.u. and py = 4 m.u. repectively. Given the budget restriction, maximize the production of bottles. By means of the Lagrange multiplier how will the maximum number of bottles produced be modified if the budget is increased in one unit (or if it is decreased)?

Answer 1

$2x+4y=68\\2x+4y-68=0..…(1)$

Production function

$q(x,y)=60x+90y-2x^2-3y^2....(2)$

Applying Lagrange multiplier method

$\frac{∆}{∆x}(60x+90y-2x^2-3y^2=\lambda\frac{∆}{∆x}(2x+4y-68)\\=60-4x=2\lambda\\=4\lambda=60-2\lambda\\x=\frac {60-2/\lambda}{4}\\=x=\frac{30-\lambda}{2}....(3)$

Partially differentiating equition 2 and 3 w.r.t y while applying Lagrange multiplier method

$\frac{∆}{∆y}(60x+90y-2x^2-3y^2=\lambda\frac{∆}{∆y}(2x+4y-68)\\=90-6y=\lambda.....(4)\\=6y=90-4\lambda=\frac{90-4\lambda}{6}\\y=\frac{45-2\lambda}{3}....(4)$

Now

$2\lambda+4y=68\\=2\frac{30-\lambda}{2}+4(\frac{45-2 \lambda}{3}=68\\=(30-\lambda)+\frac{180-8\lambda}{3}=68$

$\frac{90-3\lambda+180-8 \lambda}{3}=68\\=270-1\lambda=204\\=11\lambda=66\\\lambda=6$

$X=\frac{30-\lambda}{2}=\frac{30-6}{2}\\x=12$

$y=\frac{45-2\lambda}{3}=\frac{45-2×6}{3}=\frac{33}{3}\\=y=11$

q_max$(12,11)=60×12+90×11-2(12)^2-3(11)^2\\=105g$bottles

If budget is increased in one more unit

$2x+4y=69$

Substituting the values of x and from equition 3 and 4

$=2\frac{30-\lambda}{2}+4(\frac{45-2\lambda}{3})=69\\=3(30-\lambda)+4(25-2\lambda)=69×3\\\lambda=5=76$

$x=\frac{30-5.73}{2}=12.135\\y=\frac{45-2×5.473}{3}=11.18$

q_(x,y)$=60x+9y-2x^2-3y^2$

q _max$(12.135,11.18)=60×12.235+90×11.18-2(12.135)^2-3(11.18)^2\\=105g$

q_max$(12.135,11.18)=1064.8bottles$

The firm will produce approximately 6more bottles if budget is increased by 1 monetary unit

Incase thee budget is decreased by one unit

$2x+4y=67$

Substituting the values of x and y from equition 3 and 4

$=2\frac{30-\lambda}{2}+4(\frac{45-2\lambda}{3})=67\\=3(30-\lambda)+4(45-2\lambda)=67×3$

$=90-3\lambda+180-8\lambda=201\\=-11\lambda+270=201\\\lambda=6.27$

$x=\frac{30-6.27}{2}=11.865\\y=\frac{4.5-2×6.27}{3}=10.82$

q_(x,y)$==60x+9y-2x^2-3y^2$

$=(11.865,10.82)=60×(11.865)+9×10.82-2(11.865)^2-3(10.82)^2\\=1052.9 bottes$

Therefore approximately 6 bottles less if the budget Is decreases by 1 monetary unit.