1
ECMB02F -- Problem Set 0 Solutions
1 a) By substitution: we know that 3y = 18 - 2x, so y = 6 - (2/3)x
Thus, U(x,y) = x2y = x2[6-(2/3)x] = 6x2 - (2/3)x3 = g(x), and dg/dx = 12x - 2x2
If we set this equal to zero, we get 2x(6-x) = 0, so x = 0 or x = 6
Because g" = 12 - 4x, we see that g">0 at x=0 and g"<0 at x=6
Hence, x = 0 is a minimum and x = 6 is a maximum
Because y = 6 - (2/3)x, y = 2 when x = 6, so U(x,y) is maximized at x=6 y=2
By Lagrangian multiplier, we set L = x2y + λ(18-2x-3y)
To maximize we set
∂L/∂x = 2xy - 2λ = 0
so xy = λ
2
∂L/∂y = x - 3λ = 0
so x2 = 3λ
∂L/∂λ = 18-2x-3y = 0
Dividing the first equation by the second, we get y/x = 1/3 or 3y = x
Plugging this into the third equation, we get 18-2x-x=0, from which x = 6, and y = (1/3)x = 2
b) By substitution: we know that 2y = 18 - x, so y = 9 - (1/2)x
Thus, U(x,y) = x2y = x2[9-(1/2)x] = 9x2 - (1/2)x3 = g(x), and dg/dx = 18x - (3/2)x2
If we set this equal to zero, we get (3/2)x(12-x) = 0, so x = 0 or x = 12
Because g" = 18 - 3x, we see that g">0 at x=0 and g"<0 at x=12
Hence, x = 0 is a minimum and x = 12 is a maximum
Because y = 9 - (1/2)x, y = 3 when x = 12, so U(x,y) is maximized at x=12 y=3
By Lagrangian multiplier, we set L = x2y + λ(18-x-2y)
To maximize we set
∂L/∂x = 2xy - λ = 0 so xy = (1/2)λ
∂L/∂y = x2 - 2λ = 0 so x2 = 2λ
∂L/∂λ = 18-x-2y = 0
Dividing the first equation by the second, we get y/x = 1/4 or 4y = x
Plugging this into the third equation, we get 18-x-(½)x=0, or 18 - (3/2)x = 0, or x = 12
We use this to find y = 3
c) By substitution: we know that by = 18 - ax, so y = 18/b - (a/b)x
Thus, U(x,y) = x2y = x2[18/b-(a/b)x] = (18/b)x2 - (a/b)x3 = g(x), and dg/dx = (36/b)x - (3a/b)x2
If we set this equal to zero, we get (3/b)x(12-ax) = 0, so x = 0 or x = 12/a
Because g" = 36/b - (6a/b)x, we see that g">0 at x=0 and g"<0 at x=12/a
Hence, x = 0 is a minimum and x = 12/a is a maximum
Because y = 18/b - (a/b)x, y = 18/b - 12/b = 6/b when x = 12/a, so U(x,y) is maximized at
x=12/a, y=6/b
By Lagrangian multiplier, we set L = x2y + λ(18-ax-by)
To maximize we set
∂L/∂x = 2xy - λa = 0 so xy = (1/2)λa
∂L/∂y = x2 - λb = 0 so x2 = λb
∂L/∂λ = 18-ax-by = 0
Dividing the first equation by the second, we get y/x = a/2b or y = ax/2b
Plugging this into the third equation, we get 18-ax-b(a/2b)x=0, or 18 - ax - (a/2)x = 0
or (3a/2) x = 18 or x = 12/a. We use this to find y = (a/2b)x = (a/2b)(12/a) = 6/b
2
2 a) By substitution: we know that 3y = 18 - 2x, so y = 6 - (2/3)x
Thus, U(x,y) = x.5y.5 = x.5[6-(2/3)x].5 = [6x - (2/3)x2].5 = g(x),
and dg/dx = .5[6x - (2/3)x2]-.5[6-(4/3)x]
If we set this equal to zero, we get 6-(4/3)x = 0, so x = 6(3/4) = 4.5
Because y = 6 - (2/3)x, y = 3 when x = 4.5, so U(x,y) is maximized at x=4.5 y=3
By Lagrangian multiplier, we set L = x.5y.5 + λ(18-2x-3y)
To maximize we set
∂L/∂x = .5x-.5y.5 - 2λ = 0
so x-.5y.5= 4λ
∂L/∂y = .5x.5y-.5 - 3λ = 0
so x.5y-.5 = 6λ
∂L/∂λ = 18-2x-3y = 0
Dividing the first equation by the second, we get y/x = 2/3 or 3y = 2x
Plugging this into the third equation, we get 18-2x-2x=0, from which x = 4.5 and y = (2/3)x = 3
b) By substitution: we know that 3y = 18 - 2x, so y = 6 - (2/3)x
Thus, U(x,y) = xy = x[6-(2/3)x] = 6x - (2/3)x2 = g(x), and dg/dx = 6 - (4/3)x
If we set this equal to zero, we get 6 - (4/3)x = 0, so x = 4.5
Because y = 6 - (2/3)x, y = 3 when x = 4.5, so U(x,y) is maximized at x=4.5 y=3
By Lagrangian multiplier, we set L = xy + λ(18-2x-3y)
To maximize we set
∂L/∂x = y - 2λ = 0
so y = 2λ
∂L/∂y = x - 3λ = 0
so x = 3λ
∂L/∂λ = 18-2x-3y = 0
Dividing the first equation by the second, we get y/x = 2/3 or 3y = 2x
Plugging this into the third equation, we get 18-4x=0, from which x = 4.5, and y = (2/3)x = 3
We get the same answer in parts a and b because xy = [x.5y.5]2, and maximizing a function is the
same as maximizing its square (assuming it never goes negative, which is assured in this case
because we are taking square roots).
3a) By Lagrangian multiplier, we set L = 2L0.5 + K0.5 + λ(60-6L-K)
To maximize we set
∂L/∂L = L-0.5 - 6λ = 0
so L-0.5 = 6λ
-0.5
∂L/∂K = 0.5K - λ = 0
so K-0.5 = 2λ
∂L/∂λ = 60-6L-K = 0
Dividing the first equation by the second, we get L-0.5/K-0.5 = 3 or K0.5/L0.5 = 3 or K/L = 9
We can thus plug K = 9L into the third equation to get 60-15L=0, from which L = 4
We use this to find K = 9L = 36
b) By Lagrangian multiplier, we set L = 2L0.5 + K0.5 + λ(60-2L-K)
so L-0.5 = 2λ
To maximize we set
∂L/∂L = L-0.5 - 2λ = 0
so K-0.5 = 2λ
∂L/∂K = 0.5K-0.5 - λ = 0
∂L/∂λ = 60-2L-K = 0
Dividing the first equation by the second, we get L-0.5/K-0.5 = 1 or K0.5/L0.5 = 1 or K/L = 1
We can thus plug K = L into the third equation to get 60-3L=0, from which L = 20, and K = 20
3
c) We set L = 6L + K + λ(10 - 2L0.5 - K0.5)
To minimize we set
∂L/∂L = 6 - λL-0.5 = 0
so 6 = λL-0.5
so 1 = 0.5λK-0.5
∂L/∂K = 1 - 0.5λK-0.5 = 0
∂L/∂λ = 10 - 2L0.5 - K0.5 = 0
Dividing the first equation by the second, we get L-0.5/0.5K-0.5 = 6 or K0.5/L0.5 = 3 or K/L = 9
We can thus plug K = 9L into the third equation to get 10 - 2L0.5 - 3L0.5 = 0, from which we get
5L0.5 = 10 or L0.5 = 2 or L = 4, and K = 9L = 36
Since this gives us the same answer, it must somehow be the same problem. To see why this is
true, we will have to wait until later in the course!
4. By Lagrangian multiplier, we set L = (L-1 + K-1)-1 + λ(60-4L-K)
so (L-1 + K-1)-2L-2 = 4λ
To maximize we set ∂L/∂L = -1(L-1 + K-1)-2(-1)L-2 - 4λ = 0
-1
-1 -2
-2
∂L/∂K = -1(L + K ) (-1)K - λ = 0
so (L-1 + K-1)-2K-2 = λ
∂L/∂λ = 60-4L-K = 0
Dividing the first equation by the second, we get L-2/K-2 = 4 or K2/L2 = 4 or K/L = 2
We can thus plug K = 2L into the third equation to get 60-6L=0, from which L = 10
We use this to find K = 2L = 20
To solve the same problem by substitution, it is probably easier to simplify the function we are
trying to maximize: Q = (L-1 + K-1)-1 = (1/L + 1/K)-1 = [(K+L)/LK]-1 = LK/(K+L)
We can plug the constraint K = 60-4L into the Q equation: Q = L(60-4L)/(60-3L)
or Q = (60L - 4L2)/(60 - 3L)
To maximize this function, we set dQ/dL = 0 and we can get dQ/dL by using the quotient rule:
dQ/dL = [(60-3L)(60-8L) - (60L-4L2)(-3)]/(60-3L)2 = 0
We can multiply through by the denominator of the left hand side to get:
(60-3L)(60-8L) - (60L-4L2)(-3) = 0 or 3600 - 660L + 24L2 + 180L - 12L2 = 0
or 3600 - 480L + 12L2 = 0 or 300 - 40L + L2 = 0 or (L-10)(L-30) = 0
This tells us the L = 10 or L = 30
If L = 30, K = 60 - 4L = -60, which is an unacceptable answer, since we were only permitted
non-negative values of L and K
If L = 10, K = 60 - 4L = 20, which is an acceptable answer.
Clearly, Lagrangians are the way to go on many of these constrained maximization problems!!