We are given the equations:
<span>5 x + 11 y = 49 -->
eqtn 1</span>
<span>u = 3 x^2 + 6 y^2 -->
eqtn 2</span>
Rewrite eqtn 1 explicit to y:
11 y = 49 – 5 x
<span>y = (49 – 5x) / 11 -->
eqtn 3</span>
Substitute eqtn 3 to eqtn 2:
u = 3 x^2 + 6 [(49 – 5x) / 11]^2
u = 3 x^2 + 6 [(2401 – 490 x + 25 x^2) / 121]
u = 3 x^2 + 14406/121 – 2940x/121 + 150x^2/121
u = 4.24 x^2 – 24.3 x + 119.06
Derive then set du/dx = 0 to get the maxima:
du/dx = 8.48 x – 24.3 = 0
solving for x:
8.48 x = 24.3
x = 2.87
so y is:
y = (49 – 5x) / 11 = (49 – 5*2.87) / 11
y = 3.15
Answer:
<span>George will choose some of each commodity but more y than
x.</span>
= 220/3
0