Then, your consumer problem has both a time constraint t1+ t2+ t3= 8 and a budget constraint p1t1+ p2t2= wt3

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1082 模模模03-07班班班期期期考考考手手手寫寫寫題題題詳詳詳解解解及及及評評評分分分標標標準準準

1. (20%) As a financially independent NTU student, you spend 8 hours at school and need 8 hours of sleep every day.

This leaves you 8 hours of time to allocate between studying t₁ hours at the K-book center, singing for t₂ hours at the KTV, and teaching t₃ hours hours as a private tutor for high school kids. Suppose that the price of staying at the K-book center is p₁ dollars per hour, the price of singing at the KTV is p₂ dollars per hour, and your hourly wage as a private tutor is w dollars. Then, your consumer problem has both a time constraint t₁+ t²+ t³= 8 and a budget constraint p1t1+ p²t2= wt³. Suppose that your utility function is f(t¹, t2, t3) = ¹2ln t1+¹2ln t2.

(a) (14%) Find t^∗₁, t^∗₂ and t^∗₃ that maximizes the utility f(t¹, t2, t3). Find the maximum utility f(t^∗1, t^∗₂, t^∗₃). Write your answers in terms of p1, p2, and w.

(b) (6%) If your hourly wage w increases, find the rate of change for the maximum utility f(t^∗1, t^∗₂, t^∗₃) with respect to w. i.e. compute _∂w^∂ f(t^∗1, t^∗₂, t^∗₃). Write your answer in terms of p¹, p2, and w.

Solution:

(a) Solution 1:

Let g(t¹, t2, t3) = t¹+ t²+ t³, h(t¹, t2, t3) = p¹t1+ p²t2− wt³

We want to find the maximum value of f(t¹, t2, t3) = ¹2ln t1+¹2ln t2under constraints g= 8, h = 0.

By the method of Lagrange multipliers, we solve the system of equations

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎩

ft₁= λg^t1+ µh^t1

f_t₂= λg^t2+ µh^t2

ft₃= λg^t3+ µh^t3

g= 8 h= 0

⇒

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎩

1 2 1

t1 = λ + p¹µ⋯ ⋅ 1O

1 2 1

t2 = λ + p²µ⋯⋯ 2O 0= λ − µw⋯⋯ 3O t1+ t²+ t³= 8⋯⋯ 4O p₁t₁+ p²t₂− wt³= 0⋯⋯ 5O (5pts for correct setting and equations.)

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎨⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎩

O3 ⇒ λ = µ ⋅ w O1 ⇒ t¹= 2µ(w+p¹ ₁)

O2 ⇒ t²= 2µ(w+p¹ ₂)

O4 × w + 5O⇒ (w + p¹)t¹+ (w + p²)t²= 8w ⇒¹u= 8w.Hence t¹= w+p^4w₁, t2= w+p^4w₂

O5 ⇒ t³= w+p^4p¹1+w+p^4p²2

(7 pts for solving equations)

Hence we have only one solution(t^∗1, t^∗₂, t^∗₃) = (w+p^4w₁,_w+p^4w

2,_w+p^4p¹

1+w+p^4p²₂) with µ= 8w¹ , λ=¹8.

We know that f should obtain maximum value on the constraint set. Therefore f(t^∗1, t^∗₂, t^∗₃) is the maximum utility f(t^∗1, t^∗₂, t^∗₃) = ¹2ln(w+p^4w₁) +¹2ln(w+p^4w₂).

(2 pts for correct answers) Solution 2:

Solve⎧⎪⎪

⎨⎪⎪⎩

t1+ t²+ t³= 8

p₁t₁+ p²t₂= wt³ ⇒⎧⎪⎪

⎨⎪⎪⎩

(p²− p¹)t¹= 8p²− (w + p²)t³ (P²− P¹)t²= −8p¹+ (w + p¹)t³

⇒ t¹= p^8p2−p²1−p^w+p2−p²1t₃, t₂=p^−8p2−p¹1 +p^w+p2−p¹1t₃.

f(t¹, t₂, t₃) = 1

2ln t₁+1

2ln t₂= 1

2ln( 8p₂

p2− p¹ − w+ p² p2− p¹t₃) +1

2ln( −8p₁

p2− p¹+ w+ p¹ p2− p¹t₃) is a function of single variable t3.

Find t3 such that the above function is maximized. . . 6pts Solve the correct answer 8pts.

(b) _∂w^∂ f(t^∗1, t^∗₂, t^∗₃)¹2(w¹ −w+p¹1) +¹2(w¹ −w+p¹2) = w¹ −¹2(w+p¹1) −¹2(w+p¹2) . . . 6pts

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2. (16%) We define the improper integral over the entire plane R² I= ∬_R2e⁻¹²^(x²^+y²⁾dA as lim

a→∞∬_D

a

e⁻¹²^(x²^+y²⁾dA, where D_a is the disk with radius a and center the origin.

(a) (10%) Find∬_D

a

e⁻¹²^(x²^+y²⁾dA. Compute lim

a→∞∬_D

a

e⁻¹²^(x²^+y²⁾dA (b) (6%) The integral I can be also defined as lim

a→∞∬_S

a

e⁻¹²^(x²^+y²⁾dA, where Sa = [−a, a] × [−a, a]. From (a), Compute∫

∞

−∞

e⁻¹²^x²dx.

Solution:

(a)

∬_D

a

e⁻¹²^(x²^+y²⁾dA = ∫₀^2π∫₀^ae⁻¹²^r²rdrdθ (2 points for writing down this)

= ∫₀^2π∫_r=0^r=ae⁻¹²^r²d(1

2r²) dθ (2 points for this change of variables)

= 2π(−1)e⁻¹²^r²∣^r=a

r=0

= −2πe⁻¹²^a²+ 2π (3 points for obtaining this)

∴ lim_a→∞∬_D_ae⁻¹²^(x²^+y²⁾dA= lim_a→∞(−2πe⁻¹²^a²+ 2π) = 2π (3 points)

(b)

I = lim_a→∞∬_S

a

e⁻¹²^(x²^+y²⁾dA

= lim_a→∞∫

a

−a ∫

a

−a

e⁻¹²^(x²^+y²⁾dxdy (2 points)

= lim_a→∞∫_−a^ae⁻¹²^x²dx∫_−a^ae⁻¹²^y²dy= ∫_−∞^∞e⁻¹²^x²dx∫

∞

−∞

e⁻¹²^y²dy (2 points)

∴ ∫^−∞^∞ e⁻¹²^x²dx=√

2π (2 points)

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3. (24%) Evaluate∬_Rsin(y− x

y+ x) dA, where R is the trapezoidal region with vertices (1, 1), (2, 2),(0, 2), and (0, 4) by answering the following questions:

(a) (6%) Let u= y − x, v = y + x. Find the Jacobian ∂(x, y)

∂(u, v).

(b) (6%) Plot the region S in the uv-plane corresponding to R and label all the vertices of it.

(c) (12%) Evaluate ∬_Rsin(y− x y+ x) dA.

Solution:

(a) u= y − x, v = y + x Ô⇒ x = ¹2(v − u), y = ¹2(u + v) (2pts) Ô⇒ ∂(x, y)

∂(u, v) = ∣⁻¹²1 ¹² 2

1 2

∣ = ⁻¹2. (4pts)

(b) (1pt) The equation of the line passing through(1, 1) and (2, 2) is y − x = 0 Ô⇒ u = 0.

(1pt) The equation of the line passing through(1, 1) and (0, 2) is y + x = 2 Ô⇒ v = 2.

(1pt) The equation of the line passing through(0, 2) and (0, 4) is x = 0 Ô⇒ ¹2(v−u) = 0 Ô⇒ u−v = 0.

(1pt) The equation of the line passing through(0, 4) and (2, 2) is y + x = 4 Ô⇒ v = 4 (2pts) The vertices of the trapezoidal region are(0, 2), (2, 2), (4, 4), and (0, 4).

(c) ∬_Rsin(y− x y+ x) dA

= ∫₂⁴∫

u=v u=0

sin(u/v) ∣−1

2 ∣ du dv (3pts)

= ∫₂⁴−v

2cos(u/v)∣^u=v

u=0dv (3pts)

= ∫₂⁴v

2(1 − cos 1) dv (3pts)

= 3 (1 − cos 1). (3pts)

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