Question

5. Consider the R
2 ⠈ ’ R function f defined by
f (x, y) = ⠈ š
xy
and the R ⠈ ’ R
2
function r defined by
r (t) = ï ¿ ¾

e
2t
, cost


.
Use the General Chain Rule to determine the value of (f ⠗ ¦ r)
0
(0). [7]
6. Consider the R
2 ⠈ ’ R function f defined by
f (x, y) = 2x + 3y
2 ⠈ ’ x
2 + y
3
.
Determine how many saddle points the function f has. (Make use of the Second Order Partial Derivatives
Test for Local Extrema.) [9]
7. Use the Method of Lagrange to determine the largest possible volume that a rectangular block with a square
base can have if the height of the block plus the perimeter of its base equals 30cm

Answer 1

5)

$r'(t) = (e^t, e^t sin t + e^t cos t, e^t cos t − e^t sin t)\\ |r'(t)|= \sqrt{e^{2t}, e^{2t} (sin t + cos t)^2, e^{2t}(sin t- cos t)^2}=e^t \sqrt3\\$

Note that r(0) = (1,0,1) and r(2π) = (e^2π; 0; e^2π ). So,

$f*r=\int _0^{2 \pi}|r'(t)|dt =\int _0^{2 \pi}\sqrt{3}e^tdt =\sqrt3(e^{2 \pi}-1)$

6)

The function is given:

$g(x,y)=y^3-x^2+3y^2+x\\ g_x=-2x+1 \implies 0=-2x+1 \implies x = \frac{1}{2} ; g_y=3y^2+6y \implies 3y^2+6y=0 \implies y = 0,-2 \\$

Critical points

$(0.5,0) , (0.5,-2)$

Point of extrema

$(x,y)=(0.5,0)\\ D= -12(-2+1)=12>0\\ \implies maxima = (0.5,-2)\\ \implies saddle = (0.5,0)$

7)

$V= x^2h\\ S= x^2h+ \lambda (h+4x-30) \implies \frac{\partial }{\partial x}\left(s\right)=2xh+4 \lambda ; \frac{\partial }{\partial h}\left(s\right)=x^2+ \lambda\\ \lambda =-x^2, 2 \lambda =-xh\\ -2x^2=-xh\\ 2x=h\\ 2x+4x=30\\ x=5;h=10\\ V= x^2h=5^2*10=250 \space cubic \space units$