# Thread: practice worksheet on 2nd partials test

1. ## practice worksheet on 2nd partials test

find all the relative maxima, minima and saddle points of each of the following

a) f(x,y)=3x^2+6xy+7y^2-2x+4y
b) f(x,y)=x^3 - y^3 - 2xy+6
c) f(x,y)=4xy-x^4-y^4

problem 2) determine the dimensions of rectangular box, open at the top, having a volume of 32 cubic ft. Requiring the least amount of material for its construction.

thanks for your help....these work out problem to practice for the exams. need your help...thanks

2. Hello, samsum!

I'll do part (c) . . .

Find all the relative maxima, minima and saddle points.
. . $(c)\;f(x,y)\:=\:4xy-x^4-y^4$

Set the partial derivatives equal to zero and solve.
. . $\begin{array}{cc}f_x \:=\:4y - 4x^3\:=\:0 \quad\Rightarrow\quad y = x^3 \\ f_y \:=\:4x - 4y^3 \:=\:0\quad\Rightarrow\quad x = y^3\end{array} \begin{array}{cc}(1)\\(2)\end{array}$

Substitute (1) into (2): . $x \:=\:\left(x^3\right)^3\quad\Rightarrow\quad x^9 - x \:=\:0\quad\Rightarrow\quad x(x^8-1) \:=\:0$

. . which has roots: . $\begin{array}{ccc}x = 0,\;y = 0 \\x = 1,\;y = 1 \\ x = \text{-}1,\;y = \text{-}1\end{array}$

Second Derivative Test: . $\begin{array}{ccc}f_{xx} = -12x^2\\ f_{yy} = -12y^2 \\ f_{xy} = 4\end{array}$

Then: . $D \:=\:(f_{xx})(f_{yy}) - (f_{xy})^2 \:=\:(-12x^2)(-12y^2) - 4^2\:=\:16(9x^2y^2 - 1)$

At $(0,0):\;D \:=\:16(0 - 1) \:=\:\text{-}16$ . . . saddle point at $(0,0,0)$

At $(1,1):\;D \:=\:16(9-1) \:=\:+128$ . . . minimum at $(1,1,2)$

At $(\text{-}1,\text{-}1):\;D \:=\:16(9-1) \:=\:+128$ . . . minimum at $(\text{-}1,\text{-}1,2)$

2) Determine the dimensions of an open-top rectangular box having a volume of 32 ft³
requiring the least amount of material for its construction.

Let $x$ = length, $y$ = width, $z$ = height.

The volume is: . $xyz\,=\,32\quad\Rightarrow\quad z \,=\,\frac{32}{xy}$ [1]

The surface area is: . $A \;= \;xy + 2yz + 2xz$ [2]

Substitute [1] into [2]: . $A\;=\;xy + 2y\left(\frac{32}{xy}\right) + 2z\left(\frac{32}{xy}\right)$

. . and we have: . $A \;= \;xy + 64y^{-1} + 64x^{-1}$ . . . which we will minimize.

Set the partial derivatives equal to zero and solve.

. . $\begin{array}{cc}A_x \:= \:y - 64x^{-2} \:=\:0 \\ A_y\:=\:x-64y^{-2}\:=\:0\end{array}\quad \begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array}\quad \begin{array}{cc}y \:=\:\frac{64}{x^2}\\ xy^2\:=\:64\end{array} \begin{array}{cc}(3)\\(4)\end{array}
$

Substitute (3) into (4): . $x\left(\frac{64}{x^2}\right)^2\:=\:64\quad\Rightar row\quad x^3 = 64\quad\Rightarrow\quad\boxed{ x = 4}$

Substitute into (3): . $y \:=\:\frac{64}{4^2}\quad\Rightarrow\quad\boxed{y = 4}$

Substitute into (1): . $z \:=\:\frac{32}{4\cdot4}\quad\Rightarrow\quad\boxed {z = 2}$

Therefore, the dimensions of the box are: . $4 \times 4 \times 2$ feet.