Mathematics 41, section 1 – Multivariable Calculus Solutions for Practice Exam 3

I. A) For f(x, y) = x^{3} + y^{2} − 3x, we have

\frac{∂f}{∂x} = 3x^{2} − 3 \frac{∂f}{∂y} = 2y

The critical points are where both partial derivatives equal zero:

(x, y) = (±1, 0)

From the gradient vector field:

we see that (1, 0) is a source for ∇f and (−1, 0) is a saddle point. By the First Derivative Test (see page 293 of text), this means (1, 0) is a local minimum and (−1, 0) is a saddle point. B) The second order partial derivatives are

\frac{∂^{2}f}{∂x^{2}} = 6x  \frac{∂^{2}f}{∂x∂y} = 0  \frac{∂^{2}f}{∂y^{2}} = 2

We have

H = \frac{∂^{2}f}{∂x^{2}} \frac{∂^{2}f}{∂y^{2}} − (\frac{∂^{2}f}{∂x∂y} )^{2} = 12x − 0^{2} = 12x

So at (1, 0), H = 12 > 0 and \frac{∂^{2}f}{∂x^{2}} = 6 > 0 which shows (1, 0) is a local minimum. At (−1, 0), H = −12 < 0 so this is a saddle point. (This also agrees with A!)

C) The Lagrange Multiplier equations are:

\begin{aligned}3x^{2} − 3 & = 2λx \\ 2y & = 2λy \\ x^{2} + y^{2} & = 9 \end{aligned}

From the second equation, we get 2y(1 − λ) = 0, so either y = 0, or λ = 1.

If y = 0, then from the constraint equation, x = ±3, so we get two points (±3, 0).

If λ = 1, then from the first equation, 3x^{2} − 3 = 2x, or 3x^{2} − 2x − 3 = 0. Using the quadratic formula we get

x = \frac{2 ± \sqrt{4 − 4 · 3 · 3}}{6} = \frac{1 ± \sqrt{10}}{3} ≐ −.72, 1.39

With x = −.72, we get from the constraint equation y ≐ ±2.91; with x = 1.39, we get y ≐ ±2.66. Comparing values of f, we get

f(3, 0)  = 18 f(−3, 0) = −18
f(−.72, ±2.91) ≐ 10.25
f(1.39, ±2.66) ≐ 5.59

The maximum and minimum values are the 18, -18 respectively. D) The maximum and minimum values are the 18, -18 respectively (compare with f(1, 0) =

−2 found in part A).

II. A) Subdividing the rectangle into 9 equal smaller rectangles (so that we can use all the

data points), we have Δx = Δy = 2/3. The Riemann sum is found by adding up all the table entries times the area of the small rectangle containing the given point:

6 · \frac{2}{3} · \frac{2}{3} + 8 · \frac{2}{3} · \frac{2}{3} + ⋯ + 6 · \frac{2}{3} · \frac{2}{3} = 68/3

B)

∫_{1}^{3} ∫_{0}^{2} 2x + y^{2} dy dx = ∫_{1}^{3}(2xy + y^{3}/3)|_{0}^{2} dx
= ∫_{1}^{3} 4x + 8/3 dx
= 2x^{2} + 8x/3|_{1}^{3}
= 64/3

III. A) We must split the region at x = 0, since the lower edge is y = −x to the left of x = 0

and the lower edge is y = x to the right.

∫_{−3/\sqrt{2}}^{0} ∫_{−x}^{√\overline{9−x^{2}}} y dy dx + ∫_{0}^{3/\sqrt{2}} ∫_{x}^{√\overline{9−x^{2}}} y dy dx

B) We must also split the region here since the left and right boundaries are the lines x = −y and x = y below y = 3sqrt2 and the left and right halves of the circle above:

∫_{0}^{3/\sqrt{2}} ∫_{−y}^{y} y dx dy + ∫_{3/\sqrt{2}}^{3} ∫_{−\sqrt{9−y^{2}}}^{√\overline{9−y^{2}}} y dx dy

C) Using either form, the value is 9\sqrt{2}.