Problem 9 on the exam: Maximize the surface area of a box (with a lid) with volume

1000 $1000$ (without using Lagrange multipliers). Surface area equation:

SA=2xy+2yz+2xz $SA=2xy+2yz+2xz$ and volume equation

V=xyz=1000 $V=xyz=1000$ . (NOTE: The intention here was to force them to use the second derivative test. Recall that we only have two ways to optimize functions in Calc 3: either the second derivative test or Lagrange multipliers. Students very commonly tried a third, mysterious, method.)

WRONG solution that was very common: Let

f(x,y,z)=2xy+2yz+2xz $f(x,y,z)=2xy+2yz+2xz$ . Consider the system

⎧ ⎩ ⎨ ⎪ ⎪ f x = 2 y + 2 z = 0 f y = 2 x + 2 z = 0 f z = 2 y + 2 x = 0 (i) (i i) (i i i)

$\left\{ \begin{array}{ll} f_x = 2y+2z = 0 & (i)\\ f_y = 2x + 2z = 0 & (ii) \\ f_z = 2y + 2x = 0 & (iii) \end{array} \right.$ At this point there were two variations which I will discuss:

VARIATION 1: Set equation

(i) $(i)$ equal to equation

(ii) $(ii)$ and get

2 y + 2 z = 2 x + 2 z,

$2y+2z=2x+2z,$ therefore

(†) y = x .

$(\dagger) \hspace{35pt} y=x.$ Now set equation

(ii) $(ii)$ equal to equation

(iii) $(iii)$ and get

2 x + 2 z = 2 y + 2 x,

$2x+2z=2y+2x,$ therefore

z = y .

$z=y.$ From

(†) $(\dagger)$ we deduce that

x=y=z $x=y=z$ . Now take the volume equation, which now reads

xyz=1000 $xyz=1000$ and substitute our findings to get

x3=1000 $x^3=1000$ and thus

x=1000−−−−√3=10 $x=\sqrt[3]{1000}=10$ . Therefore both

y $y$ and

z $z$ are

10 $10$ and the max surface area is given by

SA=2(10)(10)+2(10)(10)+2(10)(10)=600 $SA=2(10)(10)+2(10)(10)+2(10)(10)=600$ . WHY THIS IS WRONG: The conclusion that

y=x $y=x$ was not checked in equation

(iii) $(iii)$ . If it were, you would find that

(iii) $(iii)$ states

2y+2x=0 $2y+2x=0$ and since

y=x $y=x$ this yields

4x=0 $4x=0$ and therefore

x=0 $x=0$ . The conclusion that

x=0 $x=0$ here is unavoidable and therefore the conclusion that

x=10 $x=10$ later is a logical contradiction.
-----------------------------
VARIATION 2: In equation

(i) $(i)$ solve for

y $y$ and see that

(‡) y = - z .

$(\ddagger) \hspace{35pt} y=-z.$ Plug this into equation

(ii) $(ii)$ to deduce that

2x−2y=0 $2x-2y=0$ and therefore

x = y .

$x=y.$ At this point most people either jumped to the

xyz=1000 $xyz=1000$ equation and concluded that

x=y=z=10 $x=y=z=10$ somehow (I saw MANY feats of mental gymnastics to justify this--all were wrong).
WHY THIS IS WRONG:: Formula

(‡) $(\ddagger)$ says

y=−z $y=-z$ but the conclusion made says

x=y=z=10 $x=y=z=10$ . It's impossible for both

y=−z $y=-z$ and

y=10=z $y=10=z$ to hold. That is a logical contradiction.
-----------------------------
ACTUAL SOLUTION: Use the volume equation to deduce that

z=1000xy $z=\dfrac{1000}{xy}$ and plug that into the surface area equation to get the function

f (x, y) = 2 x y + 2000 x + 2000 y .

$f(x,y)=2xy + \dfrac{2000}{x} + \dfrac{2000}{y}.$ Find critical points of this function by solving the system of equations

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ f x = 2 y - 2000 x 2 = 0 f y = 2 x - 2000 y 2 = 0 (A) (B)

$\left\{ \begin{array}{lll} f_x = 2y - \dfrac{2000}{x^2} = 0& (A) \\ f_y = 2x - \dfrac{2000}{y^2}= 0 & (B) \end{array} \right.$ Equation

(A) $(A)$ implies that

y=1000x2 $y = \dfrac{1000}{x^2}$ . Plug this into equation

(ii) $(ii)$ to get equation

2 x - 2000 x 4 1000 2 = 0

$2x - \dfrac{2000 x^4}{1000^2} = 0$ which has real-valued solutions

x = 0, x = 10.

$x=0, x=10.$ We throw away

x=0 $x=0$ for not making physical sense (a box can't have volume

1000 $1000$ when one of its sides are length

0 $0$ ). So we must use

x=10 $x=10$ . This implies that

y = 1000 10 2 = 1000 100 = 10,

$y=\dfrac{1000}{10^2} = \dfrac{1000}{100} = 10,$ so the only critical point of our system is

x=10 $x=10$ and

y=10 $y=10$ . We can check that this is a minimum by computing

⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ f x x = 4000 x 3 f x y = 2 f y x = 2 f y y = 4000 y 3

$\left\{ \begin{array}{ll} f_{xx} = \dfrac{4000}{x^3} \\ f_{xy} = 2 \\ f_{yx} = 2 \\ f_{yy} = \dfrac{4000}{y^3} \end{array} \right.$ and therefore at

x=10,y=10 $x=10, y=10$ ,

D = ∣ ∣ ∣ f x x (10, 10) f y x (10, 10) f x y (10, 10) f y y (10, 10) ∣ ∣ ∣ = 16 - 4 = 12 > 0

$D = \left| \begin{array}{ll} f_{xx}(10,10) & f_{xy}(10,10) \\ f_{yx}(10,10) & f_{yy}(10,10) \end{array} \right| = 16 - 4 = 12 > 0$ So since at the critical point,

D>0 $D>0$ and

fxx(10,10)>0 $f_{xx}(10,10)>0$ , we must conclude by the second derivative test that the critical point is a local minimum.

