Instructor: Shane Murphy
Econ 100 Optimization practice worksheet solutions
1. What is the derivative of y = 1 − x3 + x1/2 ?
−1
dy
Solution: dx
= 0 − 3x2 + 12 x1/2−1 = −3x2 + 12 x 2
2. What is the derivative of y = x(1 − x) + x3 − 1?
dy
Solution: y = x − x2 + x3 − 1 so dx
= 1 − 2x + 3x2 .
3. The derivative of a function is its slope. What is the slope of y = 1 − 2x? Solution:
slope is −2.
4. What is the slope of y = x(1 − x) + x3 − 1 at x = 1. Solution:
dy
2
dx at x = 1 = 1 − 2(1) + 3(1) = 1 − 2 + 3 = 2.
dy
dx
dy
dx
= −2, so the
= 1 − 2x + 3x2 , so plug in x = 1:
5.
6. Where is the slope of y = 4x − x2 equal to zero?
dy
Solution: dx
= 4 − 2x = 0 ⇒ 4 = 2x ⇒ x = 2.
7. Treat u as y and q as x. Where is the slope of u = 2 − 2q 3/2 equal to zero?
3 12
Solution: du
dq = 2( 2 )q = 0 ⇒ q = 0.
8. What is the optimum of u = x(2 − x) + 5?
Solution: u = 2x − x2 + 5 ⇒ du
dx = 2 − 2x = 0 ⇒ 2 = 2x ⇒ x = 1. A x = 1, u = 1(2 − 1) + 5 = 6.
9. Treat a as y. What is the slope of the budget constraint given by 4 = 3a + 1b?
1
Solution: solve for a, a = 43 − 13 b, so the slope is, da
db = − 3 .
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10. If u is constant, what is the slope of the indifference curve u = a + b 3 ?
1
1 −2
1 −2
3 = − b 3
Solution: solve for a, a = u − b 3 . So da
db = 0 − 3 b
3
11. For what value of b are these equal? Plug that b into the budget constraint and solve for a.
−2
−2
Solution: − 13 = − 31 b 3 ⇒ 1 = b 3 ⇒b=1 .
The budget constraint is 4 = 3a + b ⇒ 4 = 3a + 1 ⇒ 3 = 3a ⇒ a = 1.
12. Now solve the budget constraint for a and plug that into the utility function. You should get
1
u = 43 − 31 b + b 3 .
1
1
Solution: 4 = 3a + b ⇒ 3a = 4 − b ⇒ a = 43 − 13 b. Plug into u = a + b 3 and we get u = 43 − 31 b + b 3 .
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13. Where is the slope of u = 43 − 13 b + b 3 equal to zero? What is its optimum?
−2
1
1 −2
1
1 −2
1
1 −2
3 set equal to 0, 0 = − + b 3 ⇒
3 ⇒ 1 = b 3 ⇒ 1 = b. At b = 1,
Solution: du
db = − 3 + 3 b
3
3
3 = 3b
1
u = 43 − 31 (1) + (1) 3 = 34 − 13 + 1 = 33 + 1 = 1 + 1 = 2.
14. If a firm has a profit function of π = −12 + 12q − 3q 2 , what is the optimum q? What is the profit at
that q?
2
Solution: dπ
dq = 12 − 6q = 0 ⇒ q = 2. So π = −12 + 12(2) − 3(2 ) = −12 + 24 − 12 = 0.
15. If a perfectly competative firm faces a price of 12, a fixed cost of 12 and a variable cost of −3q per
unit sold (so variable cost is −3q ∗ q, what is the firms total revenue function? Total cost function?
profit function? What is the optimum q and what is the profit at that q?
Solution: T R = pq = 12q, T C = F C+V C = 12+−3q 2 so profit is π = 12q−(12+3q 2 ) = 12q−12−3q 2 ,
which is the same as the previous problem, q = 2 and π = 0..
16. If a perfectly competative firm faces a price of 4, no fixed cost, and a variable cost of −2q per unit
sold (so variable cost is −2q ∗ q, what is the firms total revenue function? Total cost function? profit
function? What is the optimum q and what is the profit at that q?
Solution: T R = pq = 4q, T C = F C + V C = 0 + −2q 2 so profit is π = 4q − 2q 2 ⇒ dπ
dq = 4 − 2(2q) =
4−4q. Set this equal to zero: 4−4q = 0 ⇒ 4 = 4q ⇒ q = 1. Plug into profit, π = 4q −2q 2 = 4−2 = 2.
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