Chapter 5 Practice Problems [Microeconomics I Studied #47]

The early problems were easy enough that I just slapped photos in and called it a day, but from here on out I’ll try to actually type things out as much as I can.

T_T T_T T_T Uuuugh, somebody do this for me, pleeeease lol lol lol

Prob 5.3

Handwritten solution sheet for Prob 5.3: Suzie buys food x and clothing y with U = xy; the tangency condition −p_x/p_y = −y/x gives xp_x = yp_y, and the budget line then gives I = 2yp_y, so the demand curve for clothing is y = I/(2p_y)

Handwritten sheet for Prob 5.3 b) and c): y = I/(2p_y) grows with I, so clothing is a normal good; then the cross elasticity of food demand with respect to the clothing price, with the note that food demand does not depend on p_y

Prob 5.7

Reggie only consumes two goods: food and housing. On a graph with housing on the x-axis and food on the y-axis, the price consumption curve for housing comes out as a vertical line. Draw a budget line and indifference curves that are consistent with these preferences.

Slide ①: food on the vertical axis, housing on the horizontal axis; say this budget line and indifference curve are the optimum

Slide ②: when housing gets cheaper, the budget line pivots out (blue) and utility rises to a higher indifference curve

Slide ③: this change in utility already includes the substitution effect from the black dot to the red dot on the old indifference curve: more housing, less food

Slide ④: the income effect (red arrow) then raises food consumption and cuts housing consumption

Slide ⑤: the final optimum (blue dot) sits straight above the old one, so the price consumption curve for housing is vertical

For reference — a good whose consumption drops because of the income effect is called an inferior good, and a good whose consumption rises from the substitution effect but then gets clobbered in the negative direction by the income effect by more than that amount is what we specifically call a GIFFEN GOOD.

Prob 5.8

Ginger’s utility function is U = x²y . So the marginal utilities are MU_x = 2xy and MU_y = x² . Income is $I = 240$ and the prices are p_x = 8 , p_y = 2 .

a) Given these prices and this income, find Ginger’s optimal basket.

Budget line: I = xp_x + yp_y, so 240 = 8x + 2y and y = −4x + 120

For U = x²y the indifference curve is dU = 0, so its slope is dy/dx = −(∂U/∂x)/(∂U/∂y) = −2xy/x² = −2y/x

At the tangency point, the slope of the budget line equals the slope of the indifference curve, so —

−4 = −2y/x gives y = 2x, the condition for equal slopes; on y = −4x + 120 that means x = 20, y = 40, and U = 20²·40 = 16000

b) If the price of y rises to 8 dollars and Ginger’s income stays the same, how low does $p_x$ have to drop for her to keep the same level of welfare she had before changed?

In one sentence, here’s the problem:

Hand sketch: the budget line before the y price rise (from 120) and after it (from 30 to 30); the arrow under the x-axis says the task is to find p_x

After the y price rise, the new budget line is y = −x + 30

When the x price drops to p_x, the new new budget line is 240 = p_x x + 8y, or y = −(p_x/8)x + 30

Slope of the new new budget line = slope of the indifference curve: −p_x/8 = −2y/x, so xp_x = 16y

Same as everything we did above: I’m going to plug this condition into the new new budget line, because we want the spot where the indifference curve of the utility function and the new new budget line are tangent.

With xp_x = 16y, the line y = −(p_x/8)x + 30 becomes y = −2y + 30, so 3y = 30 and y = 10

For utility to stay exaaactly the same, we just need the partner of $y=10$ that makes $U = 16000$.

U = x²y: the original utility was 16000 and now y = 10, so 16000 = 10x², x² = 1600, x = 40

Plug x = 40, y = 10 into y = −(p_x/8)x + 30: 10 = −5p_x + 30, so 5p_x = 20 and p_x = 4

Prob 5.11

Scott only consumes two goods, steak and ale. When the price of steak drops, he ends up buying more of both steak and ale. Explain this with budget lines and indifference curves.

Hand sketch with ale on the vertical axis and steak on the horizontal axis: ① best choice before the price drop, ② the steak price drops and the budget line pivots out, ③ best choice after the drop, ④ more steak bought, ⑤ more ale bought too

It’ll help if you follow along in numerical order. The problem just asks for an explanation, so honestly I think this one diagram is enough.

Prob 5.13

Suppose the consumer’s utility function is U = xy + 10y . (This is obvious from a partial derivative, but anyway —) MU_x = y & MU_y = x + 10 .

The price of x is p_x , the price of y is p_y , and both prices are positive. Income is .

a) For an interior optimum, show that the demand for x can be written as

Budget line I = xp_x + yp_y with slope −p_x/p_y; indifference curve U = xy + 10y with slope dy/dx = −y/(x + 10); since it’s an indifference curve, U is treated as a constant

The tangency condition (slopes equal) is

−p_x/p_y = −y/(x + 10), so yp_y = p_x(x + 10)

Plug that into the budget line —

Plugging into I = xp_x + yp_y: I = xp_x + p_x(x + 10) = 2xp_x + 10p_x, so x = I/(2p_x) − 5

QED for an interior optimum. Including the corner, demand is $x = \max\{I/(2p_x)-5,\,0\}$.

b) Now suppose $I = 100$. At what value of does consumption of x fall to zero?

Plug I = 100 into x = I/(2p_x) − 5 to get x = 50/p_x − 5; x ≥ 0 has to hold, so 50/p_x ≥ 5, p_x ≤ 10, and the max value of p_x is 10

c) Suppose the price of x is 20 and the price of y is 20 as well. On a graph showing the optimal basket of x and y, show that since exceeds the value from (b), this is a corner solution where the consumer only buys y.

The budget line with both intercepts at 5

If the prices of x and y are both 20 and income is 100, then $100 = 20x + 20y$ is the budget line, so the picture looks like this, right? Now let’s try to find the optimal point (the one that maxes out utility) with our utility function $U = xy + 10y$.

Budget line slope = −1, indifference curve slope = −y/(x + 10)

So the equal-slope condition is y/(x + 10) = 1, or y = x + 10; plug it into the budget line to find the point on the budget line that fits the condition

100 = 20x + 20y = 20x + 20(x + 10) = 40x + 200, so 40x = −100 and x = −5/2 = −2.5

The tangency equations give $x = -2.5$ and $y = 7.5$. But that point is outside the feasible region, so it is not the consumer’s optimum.

If we mark that algebraic tangency on the extended budget line above

The tangency point at x = −2.5, left of the y-axis, on the budget line

That tangency sits to the left of the y-axis. On the feasible part of the budget line, $U=(x+10)(5-x)$ falls as x increases from zero, so the best feasible choice is at the corner.

But hang on — x isn’t allowed into the negative region, $x \lt 0$.

(We said x is a good being consumed, so negative consumption is weird. It doesn’t even mean selling the good off or anything.)

So this consumer, trying to max utility “as best as possible,” settling for slightly lower utility than that, the highest utility you can get within the constraint —

The corner point x = 0, y = 5 on the y-axis (blue) with the indifference curve through it

— is going to pick $x = 0$. That’s what I’m saying.

So they pick $x = 0$, $y = 5$, and the utility from that choice is $U = xy + 10y = 0 \times 5 + 10 \times 5 = 50$.

d) Easy, skip.

e) Still assuming income is 100, draw the demand for x across all values of .

For an interior solution, the tangency condition is still the same. Across all positive prices, though, demand is $x=\max\{50/p_x-5,\,0\}$: it is zero once $p_x\geq10$. The calculation below gives the positive-demand branch —

Equal-slope condition: −p_x/p_y = −y/(x + 10), so yp_y = p_x(x + 10)

Plug it into the budget line 100 = xp_x + yp_y: 100 = 2xp_x + 10p_x, so x = 100/(2p_x) − 10p_x/(2p_x) = 50/p_x − 5

Prob 5.15

The demand for some little widget is $D = 16 - 2P$. Calculate the change in consumer surplus when the price rises from 1 dollar to 3 dollars.

First, let me sketch the function out a bit. I’m going to put $P$ on the y-axis and $D$ on the x-axis. So instead of writing $D = 16 - 2P$, it’s nicer to write $P = -(1/2)D + 8$ and draw a line with slope $-1/2$ and y-intercept 8. OK, drawing —

The demand line with P on the vertical axis and Demand on the horizontal axis

That’s what it looks like, right? We were asked for the change in consumer surplus when the price moves from 1 dollar to 3 dollars, so we need —

The same demand line with the band between P = 1 and P = 3, left of the line, shaded yellow

— the area in here, right????????

Of course, this area is easy to get with right triangles, but I’m gonna write it out the hard way for no particular reason.

Why? Just to show it can always be calculated no matter what shape the demand function takes?

The integral from 1 to 3 of (16 − 2P) dP = (−P² + 16P) from 1 to 3 = −8 + 16×2 = 24

The area is 24, and since the price went up, consumers are worse off: consumer surplus dropped by 24, so $\Delta CS = -24$.

That’s how I’d write it.

Prob 5.20

Two consumers in the market. (Jim & Donna.) Jim’s utility function is $U = xy$, Donna’s is $U = x^2 y$. Jim’s income is I_J = 100 and Donna’s income is I_D = 150 .

a) When the price of y is 1 and the price of x is P, find Jim’s and Donna’s optimal baskets.

Jim: budget line 100 = xP + y with slope −P; indifference curve U = xy with slope −y/x; equal slopes y/x = P, so y = Px

The equal-slope point has to be on the budget line: 100 = Px + Px = 2Px, so x = 50/P and y = 50; Jim’s optimal basket is (50/P, 50)

Donna: budget line 150 = xP + y with slope −P; indifference curve U = x²y with slope −2y/x; equal slopes 2y/x = P, so 2y = Px

On the budget line: 150 = Px + Px/2 = (3/2)Px, so x = 100/P and y = 50; Donna’s optimal basket is (100/P, 50)

Jim’s demand curve x_J = 50/P and Donna’s x_D = 100/P; total demand D_tot = 50/P + 100/P = 150/P, and with one more Donna, 50/P + 100/P + 100/P = 250/P

I’m pretty sure this solves everything up through (d)??????

Prob 5.22

Suppose Bart and Homer are the only people in Springfield who drink the soft drink Seven-Up. And suppose their inverse demand curves for Seven-Up are P = 10 − 4Q_B and P = 25 − 2Q_H , respectively, and obviously they don’t consume negative quantities. Draw the market demand curve for Seven-Up in Springfield.

Oh, this is just asking us to draw the two functions and then draw a new function that’s their sum.

Hand sketch: Homer’s demand P = 25 − 2Q_H and Bart’s P = 10 − 4Q_B on one graph, each solved for its quantity, and the market demand Q_tot = Q_B + Q_H written piece by piece for the two price ranges

Easy, let’s bang it out and move on. We just add them range by range, and the ranges split up naturally because the two curves hit zero at different prices: above P = 10, Bart drops out. So at the boundary you’ll get a kink. heh heh heh

Prob 5.17

Lou’s preferences over pizza (x) and other goods (y) can be written as $U(x,y) = xy$, and his income is 120.

a) When $p_x = 4$ and $p_y = 1$, find the optimal basket.

Ahh, I’m getting kind of sick of these by now….

Budget line slope = indifference curve slope: −4 = −y/x, so y = 4x

That’s the condition for the point where the slopes are equal, and now we plug that into the budget line again. Let’s plug.

I didn’t bother writing the budget line out above and just said the slope was $-4$ — because I was going to write it right here. That’s why I skipped it earlier.

Anyway, writing it now —

I = xp_x + yp_y, so 120 = 4x + y; plugging y = 4x in gives 120 = 8x, so x = 15 and y = 60, and this guy’s utility is U = xy = 15×60 = 900

Now let me draw the situation up to this point.

Lou’s optimum at (15, 60): the budget line from 120 to 30 touching the indifference curve U_1 = 900

And then we move on to part (b). The reason I drew that picture is so part (b) goes nice and smooth~.

b) When the pizza price drops to 3 dollars, calculate the income and substitution effects.

That’s the problem — but on top of the income and substitution effects, let’s go ahead and grab the equivalent variation and compensating variation while we’re at it. All of it.

First, since the price moved from 4 dollars to 3 dollars, the budget line gets drawn in red like this.

The new budget line BL_2 (red) from 120 to 40

Since $120 = 3x + y$, the slope of $BL_2$ is $-3$.

OK, let’s first find the compensating variation. Keep the slope of $BL_2$, and call the income the unknown i_2 .

A line with BL_2’s slope tangent to U_1, with the unknown income i_2 on the y-axis; the red double arrow between 120 and i_2 is the compensating variation

The amount marked by the red double-headed arrow is going to be the “compensating variation” — compensating variation: the key was “assuming the changed price,” right. I’m going to find the size of that arrow.

The line i_2 = 3x + y has slope −3 and U_1 = 900 = xy has slope −y/x; equal slopes give y = 3x, and plugging into i_2 = 3x + y gives i_2 = 6x

I need to find $x$, and I’ll get it from the $U_1$ indifference curve. Because right now $i_2$ is unknown and $U_1$ is known, instead of plugging the condition into the budget line like we’ve been doing all along, we plug it into the utility function.

(And conversely — the reason we’ve been so casually plugging into the budget line instead of the indifference curve up until now was that income was known and utility was unknown.)

So on the $U_1$ indifference curve, let’s find the $x$ where the slope is $-3$.

900 = xy with y = 3x: 900 = 3x², x² = 300, x = √300 = 10√3 (and actually y = 30√3 too)

Plugging that back into the equation above —

So the compensating variation is 120 − 60√3 , I guess. Let me mark all this on the diagram and keep going.

The compensated point (10√3, 30√3) on U_1, with i_2 marked below 120

Still with me?

The red double arrow on the x-axis between 15 and 10√3

Ah ah ah ah and the red arrow I marked in the figure above — that amount is exactly the $\Delta x$ from the substitution effect!?!?!?!?

That is,

I guess.

Now let’s go find $U_2$.

U_2 (blue dashes), the indifference curve that touches BL_2

We’re going to find a $U_2$ that looks like this — the $U_2$ that’s tangent to $BL_2$ and gives max utility!!!!!!!!!!!!!!!!!!!

BL_2: 120 = 3x + y with slope −3; U_2 = xy with slope −y/x; equal slopes y = 3x; plugging into 120 = 3x + y gives 120 = 6x, so x = 20 (and y = 60)

Whoa whoa whoa. So then $U_2 = 1200$ is what we calculate. Then back to the diagram, go go go go go. Let me mark the info we’ve found again.

The new optimum (20, 60) where BL_2 touches U_2 = 1200

Looks like the income effect is basically game over now. Earlier we found the substitution effect Δx = 10√3 − 15 , and from there, the chunk that brought us up to $x = 20$ right now is exactly the income effect. So —

The blue double arrow on the x-axis between 10√3 and 20

That amount right there is the income effect, is what I’m saying. So

Now finally all that’s left is the equivalent variation, right????????

Equivalent variation……….. The key with equivalent variation was “assuming the price before the change.” Let’s keep going with the diagram.

BL_3 (blue), with the old slope, tangent to U_2; the arrow between 120 and i_3 on the y-axis is the equivalent variation

Assuming the price before the change, we need to find the unknown $i_3$. So I named the budget line with $i_3$ as $BL_3$. “$BL_3$ — the one with a tangency point like that.” Then on with the game of hunting down $i_3$.

BL_3: i_3 = 4x + y with slope −4; U_2 = 1200 = xy with slope −y/x; equal slopes y = 4x; plugging into 1200 = xy gives 1200 = 4x², x² = 300, x = 10√3 (and y = 40√3)

BL_3 touching U_2 at (10√3, 40√3), next to the new optimum (20, 60)

Plugging into BL_3: i_3 = 4x + y = 80√3

So the equivalent variation is 80√3 − 120 , I guess.

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

In Prob 5.3 c), the food demand $x = \frac{I}{2p_x}$ has no $p_y$ in it, so the cross elasticity of food demand with respect to the clothing price is 0, not undefined.
In Prob 5.15, the demand line $D = 16 - 2P$ meets the price axis at $P = 8$ and the quantity axis at $D = 16$; the sketches label them 4 and 8. The shaded band between $P = 1$ and $P = 3$ and its area, 24 (the fall in consumer surplus), are right.
In Prob 5.22, Bart’s demand is $Q_B = 2.5 - \frac{1}{4}P$ (from $4Q_B = 10 - P$), not $10 - \frac{1}{4}P$, so for $0 \lt P \lt 10$ the market demand is $Q_H + Q_B = 15 - \frac{3}{4}P$, not $22.5 - \frac{3}{4}P$; the branch for $10 \lt P \lt 25$, $Q_H = 12.5 - \frac{1}{2}P$, is right.
In Prob 5.13, the tangency formula gives only the interior branch. Including the corner, $x=\max\{I/(2p_x)-5,0\}$. With income 100, x is positive below $p_x=10$ and zero at or above 10; 10 is not an upper bound on the price itself.