Download Answers 1. A firm`s inverse demand function is D(q)

Mathematics Practice Problems – Answers
1. A firm’s inverse demand function is D(q) = 100 − 2q. [Reminder: For each price, the demand
function gives the quantity demanded at that price. For each quantity, the inverse demand function
gives the price at which that quantity would be demanded.] Thus if a firm sells q units of the good,
it receives a price 100 − 2q for each unit sold. The firm’s total revenue is the quantity times the
price or
R(q) = 100q − 2q 2 .
a. Question: The firm’s marginal revenue is the derivative of the revenue function. Compute
the marginal revenue function. Answer: M R(q) = R0 (q) = 100 − 4q.
b. Question: Graph the firm’s demand function and the firm’s marginal revenue function.
The quantity is measured on the horizontal axis; the price and marginal revenue are measured on
the vertical axis. Answer: Both the demand and the MR functions are straight lines. For both
the demand and the MR function, the intercept on the vertical axis is 100. The intercept on the
quantity axis is 50 for the demand function and 25 for the marginal revenue function.
2. (Calculus review) A firm’s inverse demand function is D(q) = a − bq where a and b are both
positive numbers. The firm’s revenue is R(q) = aq − bq 2 .
a. Question: Compute the marginal revenue function. Answer: M R(a) = a − 2bq..
b. Question: Compare the intercept of the demand and marginal revenue functions. [The
intercept is the value of the function for q = 0.] Also compare the slopes of the demand and the
marginal revenue functions. Answer: For both the demand and the marginal revenue functions,
the intercept is a. The slope of the demand function is −b and the slope of the marginal revenue
function is −2b.
3. Question: A firm has costs C(q) = 10q 2 . Answer: The derivative of the function C is
C 0 (q) = 20q.
a. Question: Determine the value of q for which 20 = C 0 (q). Answer: Solve for the value of q for
which 20 = 20q. The solution is q = 1.
Question: Denote this value of q as q ∗ Compute 20q ∗ −C(q ∗ ). Answer: 20q ∗ −C(q ∗ ) = 20∗1−10 =
10.
b. Question: Let p be a positive price. Determine the value of q for which p = C 0 (q). [Hint:
Compute C 0 (q). Equate p to C 0 (q). Solve for q as dependent on p.] . Answer: We want to
determine the value of q for which p = C 0 (q) or p = 20q or q = p/20.
Question: Let q ∗ denote the value of q for which p = C 0 (q). Is q ∗ increasing or decreasing in p?
Answer: Since q ∗ = p/20, it is an increasing function of p.
4. Question: Solve the following two equalities for K and L in terms of w, r, and q. Your
expression for K should involve w, r, and q but not L. Similarly, your expression for L should
involve w, r, and q but not K.
w
K
=
L
r
q = KL.
Answer: There are many ways to solve two equations for two unknowns. One possible approach
is shown below.
Solve the first equation for K to obtain
wL
.
r
K=
Substitute this expression for K into the second equation to obtain
q=
wL
L.
r
Rewrite the term on the right hand side of the equality as
q=
wL2
.
r
We want to solve for L. By the above equality, L2 = rq/w. Thus,
L=(
Since we determined that K =
rq .5
) = r.5 q .5 w−.5 .
w
wL
r ,
K=
w.5 q .5
wr.5 q .5 w−.5
=
.
r
r.5
5. Question: Solve the following two equalities for K and L in terms of w, r, and q. Your
expression for K should involve w, r, and q but not L. Similarly, your expression for L should
involve w, r, and q but not K.
.5K −.5
w
=
−.5
.5L
r
q = K .5 + L.5 .
Answer: When faced with a question involving exponents, some of us have a tendency to panic
or decide that the problem is overly difficult. But, all we have to do is use our basic high-school
algebra and proceed step by step. We may require several steps in order to finally solve for K and
L but the math is familiar. We just need to be consistent in our use of algebraic principles.
We can rewrite the first equality of the problem as
w
L.5
=
K .5
r
which is also equivalent to
L.5 =
w .5
K .
r
Substitute for L.5 in the second equation to obtain
q = K .5 +
w .5
w
K = K .5 (1 + )
r
r
2
or
q/(1 +
w
) = K .5 .
r
Squaring both sides of the equality,
K = q 2 /(1 +
w 2
) .
r
Since
L.5 =
w .5
K ,
r
L = (w/r)2 K.
We have an expression for K in terms of w, r, and q. Substitute in that expression for K in order
to find L in terms of w, r, and q.
2
w 2
r
w 2
L = ( ) q 2 /(1 + ) = q 2 /( + 1) .
r
r
w
6. Question: Solve the following two equalities for x and y. Your expression for x may contain
the variables α and I but not y. Similarly, your expression for y may involve α and I but not x.
y
=2
x+a
I = 4x + 2y.
Answer: The first equality implies that y = 2(x + a). Substitute this expression for y into the
second equality to obtain
I = 4x + 2 ∗ 2(x + a).
Solving for x yields,
x=
Since y = 2(x + a),
·
y =2∗
which equals
I
4
I − 4a
.
8
¸
I − 4a
+a ,
8
+ a.
7. Question: Solve the following two equalities for x and y in terms of px , py , and I. Your
expression for x may involve px , py , and I but not y. Similarly, your expression for y may involve
px , py , and I but not x.
px
y
=
x+a
py
I = px x + py y.
Answer: The first equality implies that
y=
px
(x + a).
py
3
Substitute this expression for y into the second equality to obtain
I = px x + py ∗
Solving for x yields,
x=
Since y = 2(x + a),
px
(x + a).
py
I − px a
.
2px
·
¸
I − px a
y =2∗
+a ,
2px
which equals
I + px a
.
2py
8. The first derivative of the profit function is
Π0 (q) = 200 − 2q − 100 = 100 − 2q and the second derivative is Π00 (q) = −2.
The value of q for which the first derivative is zero is q = 50. Since the second derivative is
everywhere negative, q = 50 is a maximum. The firm’s profit at the maximum is (200 − 50) ∗ 50 −
100 ∗ 50 = 150 ∗ 50 − 100 ∗ 50 = 2500.
9. The first derivative of the profit function is
Π0 (q) = 200 − 2q − c = 200 − 2q − c and the second derivative is Π00 (q) = −2.
The value of q for which the first derivative is zero is q = 100 − c/2. Since the second derivative
is everywhere negative, q = 100 − c/2 is a maximum. To find the firm’s profit at the maximum,
substitute the quantity q = 100 − c/2 into the profit equation. The firm’s profit at the maximum
is thus
(200 − 100 + c/2) ∗ (100 − c/2) − c ∗ (100 − c/2).
Simply the first term in parentheses to obtain
(100 + c/2) ∗ (100 − c/2) − c ∗ (100 − c/2).
Note that (100 − c/2) appears in both terms of the above expression. Thus, the above expression
may be written as
(100 + c/2 − c) ∗ (100 − c/2).
Simplifying the first term yields,
(100 − c/2) ∗ (100 − c/2) = (100 − c/2)2
as the firm’s profit.
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