


Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Material Type: Exam; Class: Calculus III--Multivariable; Subject: Mathematics; University: Colgate University; Term: Spring 2004;
Typology: Exams
1 / 4
This page cannot be seen from the preview
Don't miss anything!



Math 113 – Calculus III Exam 4 Practice Problems Spring 2004
(a) Find and classify the critical points of f. (b) Find the maximum and minimum values of f subject to the constraint
x^2 + y^2 = 18
(c) Find the maximum and minimum values of f subject to the constraint
x^2 + y^2 ≤ 18
(d) Approximate the maximum value of f subject to the constraint
x^2 + y^2 = 18. 3
(Explain your answer in terms of Lagrange multipliers.)
(a)
0
∫ (^9) −y 2
0
f (x, y) dx dy
(b)
− 1
x^3
f (x, y) dy dx
(a)
0
∫ √ 16 −x 2
x
3 x dy dx
(b)
0
∫ (^) y
−y
(x^2 + 2y) dx dy
(d) The Lagrange multiplier λ gives the rate of change of the maximum value with respect to changes in the constraint constant. We can use this to approximate the change in the maximum value. Recall from (b) that at (− 3 , −3), we found λ = 5/3. The approximate change in the maximum value is then λ(18. 3 − 18) = 0 .5. Thus, the approximate maximum value of f when the constraint equation is x^2 + y^2 = 18.3 is 58.5.
(a)
0
∫ √ 9 −x
0
f (x, y) dy dx
(b)
− 1
∫ (^) y 1 / 3
− 1
f (x, y) dx dy
(a) ∫ 2 √ 2
0
∫ √ 16 −x 2
x
3 x dy dx =
0
3 xy|
√ 16 −x 2 x dx
0
3 x
16 − x^2 − 3 x^2 dx
= −(16 − x^2 )^3 /^2 − x^3
√ 2 0 = − 32
(b) ∫ (^1)
0
∫ (^) y
−y
(x^2 + 2y) dx dy =
0
x^3 3
y
−y
dy
0
2 y^3 3
y^4 6
4 y^3 3
1
∫ (^) a
−a
∫ √a (^2) −x 2
− √ a^2 −x^2
∫ (^) a−x
0
(5 + z + x^2 ) dz dy dx.