← 6. Production

Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Easy

1EasyCompute

For the production function q=z11/3z22/3q = z_1^{1/3} z_2^{2/3}, compute the MRTS at (z1,z2)=(2,8)(z_1, z_2) = (2, 8).

MRTS=αβz2z1MRTS = \frac{\alpha}{\beta}\frac{z_2}{z_1}
α=1/3\alpha = 1/3, β=2/3\beta = 2/3, reference (2,8)(2, 8).
Show solution

  1. Plug into the formula: MRTS=1/32/3⋅82=12⋅4=2MRTS = \frac{1/3}{2/3} \cdot \frac{8}{2} = \frac{1}{2} \cdot 4 = 2.

  2. So the isoquant slope is −2-2: cutting z1z_1 by one unit needs about two more units of z2z_2 to hold output constant.

Answer

MRTS=2MRTS = 2

Medium

2Medium

With w1=3w_1 = 3, w2=1w_2 = 1, q=10q = 10, α=β=1/2\alpha = \beta = 1/2, find the conditional factor demands z(w,q)z(w, q) and the minimum cost c(w,q)c(w, q).

z1=[q(αw2βw1)β]1/(α+β),z2=z1βw1αw2z_1 = \left[q\left(\frac{\alpha w_2}{\beta w_1}\right)^{\beta}\right]^{1/(\alpha+\beta)}, \quad z_2 = z_1\frac{\beta w_1}{\alpha w_2}
w=(3,1)w = (3, 1), q=10q = 10 — the tangency economizing on dear z1z_1.
Show solution

  1. z1=[10⋅(1/3)1/2]1=10/3≈5.774z_1 = [10 \cdot (1/3)^{1/2}]^{1} = 10/\sqrt{3} \approx 5.774 — economizing on the dearer z1z_1.

  2. z2=3z1=103≈17.321z_2 = 3 z_1 = 10\sqrt{3} \approx 17.321 and c=3⋅10/3+103=203≈34.64c = 3 \cdot 10/\sqrt{3} + 10\sqrt{3} = 20\sqrt{3} \approx 34.64. Check: 5.774×17.321=10\sqrt{5.774 \times 17.321} = 10 ✓.

3Medium

For C(q)=40+30q−4q2+12q3C(q) = 40 + 30q - 4q^2 + \tfrac{1}{2}q^3, find the output minimizing average variable cost and the marginal cost there.

AVC(q)=a+bq+cq2AVC(q) = a + bq + cq^2
The AVC minimum at q=4q = 4 crossed by MC.
Show solution

  1. AVC(q)=(C(q)−40)/q=30−4q+0.5q2AVC(q) = (C(q) - 40)/q = 30 - 4q + 0.5q^2. From AVC′(q)=−4+q=0AVC'(q) = -4 + q = 0, q=4q = 4.

  2. AVC(4)=30−16+8=22AVC(4) = 30 - 16 + 8 = 22, and with MC(q)=30−8q+1.5q2MC(q) = 30 - 8q + 1.5q^2, MC(4)=30−32+24=22MC(4) = 30 - 32 + 24 = 22 — the MC curve passes exactly through the AVC minimum.

Hard

4Hard

Let α=0.3\alpha = 0.3, β=0.4\beta = 0.4 (decreasing returns), w1=w2=1w_1 = w_2 = 1. Compute c(w,16)/c(w,8)c(w, 16)/c(w, 8) from the conditional-demand formula, explain why it exceeds 22, and state the degree of homogeneity of c(w,q)c(w, q) in qq in general.

c(w,tq)=t1/(α+β) c(w,q)c(w, tq) = t^{1/(\alpha+\beta)}\, c(w, q)
α=0.3\alpha = 0.3, β=0.4\beta = 0.4 (decreasing), q=8q = 8.
Show solution

  1. With s=α+β=0.7s = \alpha + \beta = 0.7, z1(w,q)=[q⋅K]1/sz_1(w, q) = [q \cdot K]^{1/s} (where K=(αw2/βw1)βK = (\alpha w_2/\beta w_1)^\beta is independent of qq), so z(w,q)z(w, q) is homogeneous of degree 1/s1/s in qq — and so is c(w,q)=w⋅z(w,q)c(w, q) = w \cdot z(w, q).

  2. Hence c(w,16)/c(w,8)=21/0.7≈2.69c(w, 16)/c(w, 8) = 2^{1/0.7} \approx 2.69. Under decreasing returns, doubling output needs more than double the inputs, so cost more than doubles. In general c(w,q)c(w, q) is homogeneous of degree 1/(α+β)1/(\alpha+\beta) in qq.