Area to Length?

Let $a, b, c$ be the side lengths of the prism. Since the areas of the faces are different, they cannot be opposite one another. So WLOG, we can write

$\begin{aligned} ab &= 20 \\ bc &= 18. \end{aligned}$

From the second equation, we have $b = \dfrac{18}{c}.$ Plugging this into the first equation yields $18a = 20c,$ or $a = \dfrac{10}{9}c.$ This eliminates the variable $a,$ and the two conditions become one condition: $bc = 18.$

The length of the space diagonal of the prism is equal to

$\sqrt{a^2 + b^2 + c^2} = \sqrt{ \left ( \dfrac{10}{9} c \right )^2 + b^2 + c^2} = \sqrt{b^2 + \dfrac{181}{81}c^2}.$

By the AM-GM inequality,

$\begin{aligned} b^2 + \dfrac{181}{81}c^2 &\geq 2\sqrt{b^2 \left ( \dfrac{181}{81}c^2 \right )} \\ &= 2bc \left ( \dfrac{\sqrt{181}}{9} \right ) \\ &= 2(18)\left ( \dfrac{\sqrt{181}}{9} \right ) \\ &= 4 \sqrt{181}. \end{aligned}$

Therefore, the minimum possible length of the space diagonal is $\sqrt{4 \sqrt{181}} \approx \boxed{7.3358}.$ Equality holds when $b^2 = \dfrac{181}{81}c^2,$ so that

$(a, b, c) = \left ( \dfrac{10\sqrt{2}}{\sqrt[4]{181}}, \sqrt{2} \cdot \sqrt[4]{181}, \dfrac{9\sqrt{2}}{\sqrt[4]{181}} \right ) \approx (3.857, 5.187, 3.470).$

EDIT: I just realized there was a more direct way to apply AM-GM to solve the problem:

$\begin{aligned} a^2 + b^2 + c^2 &\geq 2\sqrt{b^2(a^2 + c^2)} \\ &= 2\sqrt{(ab)^2 + (bc)^2} \\ &= 2\sqrt{20^2 + 18^2} \\ &= 4 \sqrt{181}. \end{aligned}$

Thus, the minimum possible length of the space diagonal is $\sqrt{4 \sqrt{181}} = 2 \sqrt[4]{181}.$ This method can also be generalized: if two faces of a rectangular cuboid have areas $A$ and $B,$ where $A \neq B,$ then the minimum possible length of its space diagonal is $\sqrt{2\sqrt{A^2 + B^2}}.$

We have the two conditions $ab = 20$ and $bc = 18$ . Multiply the second equation by $a$ to get $abc = 18a \implies 20c = 18a \implies c = \frac{9a}{10}$ . We also have, from the first equation, that $a = \frac{20}{b}$ .

With this information, we want to minimize $d = \sqrt{a^2 + b^2 + c^2}$ which is actually equivalent to minimizing $d^2 = a^2 + b^2 + c^2$ and this is much easier to differentiate. Plugging in the previous equations we have that $d^2 = \frac{b^4 + 724}{b^2}$ and the derivative with respect to $b$ of this is $2b - 1448b^{-3}$ and setting this equal to 0 we get $b= \sqrt{2} \sqrt[4]{181}$ .

Plugging this into our expression for $d$ we find that $d^2 \approx 53.8145$ and thus $d \approx 7.33584$ .

We minimize the function $f(a,b,c) = a^2 + b^2 + c^2$ under the conditions $ab = 20, ac = 18$ .

The total derivatives of these expressions are $\begin{cases} 2a\:da + 2b\:db + 2c\:dc = 0 \\ a\:db + b\:da = 0 \\ a\:dc + c\:da = 0 \end{cases}$ Multiply the first equation by $a/2$ , the second by $-b$ , the third by $-c$ ; then add: $\begin{array}{rrrl} a^2\:da & +\ ab\:db & +\ ac\:dc & = 0 \\ -\ b^2\:da & -\ ab\:db & & = 0 \\ -\ c^2\:da & & -\ ac\:dc & = 0 \\ \hline (a^2 -b^2 - c^2)\:da & & & = 0.\end{array}$ Thus the solution is characterized by the fact that $a^2 = b^2 + c^2$ . Multiply by $a^2$ : $a^4 - (ab)^2 - (ac)^2 = 0.$ Substitute the known values for $ab$ and $bc$ to obtain $a^2 = \sqrt{20^2 + 18^2}$ . The fact that $b^2 + c^2 = a^2$ provides a shortcut to find the body diagonal $D$ : $D^2 = a^2 + (b^2 + c^2) = 2a^2 = 2\sqrt{20^2 + 18^2} = 4\sqrt{181},$ $D = \sqrt{a^2 + b^2 + c^2} = \boxed{2\sqrt[4]{181} \approx 7.336}.$

Relevant wiki: Lagrange Multipliers

Since I literally just read the short section on Lagrange Multipliers in my calculus textbook, I thought I'd try applying it to this question. (The passage is much more basic than the related wiki on this site, so my notation might be a bit on the simplistic side.)

We are effectively trying to minimize $d^{2}=f(x,y,z)=x^{2}+y^{2}+z^{2}$ , subject to constraints $g(x,y,z)=xy=18$ and $h(x,y,z)=xz=20.$

The Lagrange condition is $\nabla{f}=\lambda\nabla{g}+\mu\nabla{h}$ which written in terms of its components would be $\begin{aligned} f_{x} &= \lambda g_{x}+\mu h_{x}\\ f_{y} &= \lambda g_{y}+\mu h_{y}\\ f_{z} &= \lambda g_{z}+\mu h_{z}\\ \end{aligned}$ [Note: $f_{x},f_{y}$ and $f_{z}$ are the partial derivatives w.r.t. $x,y$ and $z$ respectively.]

Applying this to our functions, and combining with our constraints gives us the system \begin{aligned} 2x&=\lambda y+\mu z \tag{1}\\ 2y&=\lambda x \tag{2}\\ 2z&=\mu x \tag{3}\\ xy&=18 \tag{4}\\ xz&=20 \tag{5}\\ \end{aligned} Multiplying equation (1) by $x$ , equation (2) by $y$ and equation (3) by $z$ yields $\begin{aligned} 2x^{2}&=\lambda xy+\mu xz\\ 2y^{2}&=\lambda xy\\ 2z^{2}&=\mu xz\\ \end{aligned}$ which immediately leads to $x^{2}=y^{2}+z^{2}$ . From there we proceed as follows: $\begin{aligned} x^{4}&=x^{2}y^{2}+x^{2}z^{2} &&\qquad (\text{multiplying by }x^{2})\\ x^{4}&=18^{2}+20^{2} &&\qquad (\text{substituting our constraints})\\ x^{2}&=2\sqrt{181}\\ \end{aligned}$ This is all we need to calculate the length of the diagonal. $\begin{aligned} d&=\sqrt{x^{2}+y^{2}+z^{2}}\\ d&=\sqrt{2x^{2}} \qquad \qquad (\text{since }x^{2}=y^{2}+z^{2})\\ d&=2\sqrt[4]{181}\\ d&\approx 7.336\\ \end{aligned}$ Strictly speaking, the Lagrange Multiplier method will identify maximum and/or minimum values, $\textit{assuming these extreme values exist}$ . So the above shows that if the space diagonal in question has a minimum length, it would be $\approx 7.336$ .

$xy = 20$ and $zy = 18 \implies y =\dfrac{20}{x} \implies z = \dfrac{9x}{10}$

$\implies D(x) = d^2(x) = \dfrac{181}{100} x^2 + \dfrac{400}{x^2} \implies \dfrac{dD}{dx} = \dfrac{181}{50} x - \dfrac{800}{x^3} = \dfrac{181 x^4 - 800 * 50}{50x^3} = 0 \implies$

$x^4 = \dfrac{10^4 * 4}{181} \implies x = \dfrac{10\sqrt{2}}{181^{1/4}}$ for $x > 0$ .

Testing for relative min using second derivative: $\dfrac{d^2D}{dx^2} = \dfrac{180}{50} - \dfrac{2400}{x^4}$

$x = \dfrac{10\sqrt{2}}{181^{1/4}} \implies \dfrac{d^2D}{dx^2} = \dfrac{-181}{25} < 0 \implies$ min at $x = \dfrac{10\sqrt{2}}{181^{1/4}}$ .

$\therefore d^2 = 4\sqrt{181} \implies d = 2\sqrt[4]{181} \approx \boxed{7.3358}$

Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality

Let the dimensions of this cuboid be $a,b,c$ such that $ab = 20, ac = 18$ . We want to find the minimum value of $d: = \sqrt{a^2 + b^2 + c^2}$ .

From $ab =20 , ac = 18$ , we get $b = \dfrac{20}a , c = \dfrac{18}a$ . Thus, $a^2 + b^2 + c^2 = a^2 + \left( \dfrac{20}a \right)^2 + \left( \dfrac{18}a\right)^2 = a^2 + \dfrac{724}{a^2}$ .

By arithmetic mean geometric mean inequality , $AM\left( a^2 , \dfrac{724}{a^2} \right) \geq GM \left( a^2 , \dfrac{724}{a^2} \right) \quad \Leftrightarrow \quad a^2 + \dfrac{724}{a^2} \geq 2 \sqrt{ a^2 \cdot \dfrac{724}{a^2} } = 2 \sqrt{ 724} .$

Thus, the minimum value of $d$ is $\sqrt{2 \sqrt{724}} = 2 \sqrt[4]{181} \approx \boxed{7.336 }$ and it occurs when $a^2 = \dfrac{724}{a^2}$ , or equivalently, when its dimensions are $\sqrt[4]{724} , \dfrac{20}{\sqrt[4]{724}} , \dfrac{18}{\sqrt[4]{724}}$ .

The rectangular cuboid have sides $a, b, c$ and the conditions are: $ac=20$ and $bc=18$ . The diagonal $d$ is going to be $d^2=a^2+b^2+c^2$ .

From this we have: $a=$ $\frac{20}{c}$ $b=$ $\frac{18}{c}$

We can plug this into the equation for the diagonal:

$d^2=(\frac{20}{c})^2 + (\frac{18}{c})^2 + c^2$

Considering only the real and positive solution:

$d=\frac{ \sqrt{724+c^4} }{c}$

With this formula for the diagonal we can find the minima by looking at the derivative and making it equal to zero.

$(\frac{ \sqrt{724+c^4} }{c})' = 0$

By using the chain rule and the division rule we get:

$\frac{ \frac{1}{2} (724+c^4)^{- \frac{1}{2}} 4 c^3 c - 1 \sqrt{724 +c^4} }{c^2} = 0$

After some algebra we get that the minimum value for $c$ is $\sqrt[4]{724}$ .

Plugging the value into the original formula:

$d=\frac{ \sqrt{724+c^4} }{c}$

$d=\frac{ \sqrt{724+ \sqrt[4]{724}^4} }{ \sqrt[4]{724} }$

And we get the minimum value for $d$ :

$d=2 \sqrt[4]{181} \approx 7.3358$

Let $x$ be the length of the edge joining the 20 and 18 faces. Then the dimensions of the cuboid are $x$ , $\frac{20}{x}$ and $\frac{18}{x}$ .

So the length of the diagonal is

$d=\sqrt{x^{2}+\left ( \frac{20}{x} \right )^{2}+\left ( \frac{18}{x} \right )^{2}}$

This is minimised when $d^{2}=x^{2}+\left ( \frac{20}{x} \right )^{2}+\left ( \frac{18}{x} \right )^{2}=x^{2}+ \frac{724}{x^{2}}$ is minimised

With a little simple calculus, solve $\frac{\mathrm{d} d^{2}}{\mathrm{d} x}=2x-2\frac{724}{x^{3}}=0$ or $x=\sqrt[4]{724}=5.187$

Substituting back into the first equation $d=\sqrt{\sqrt{724}+\frac{724}{\sqrt{724}}}=\sqrt{2\sqrt{724}}=\sqrt{4\sqrt{181}}=2 \sqrt[4]{181}=7.3358$

let a = 20k then b = 1/k (ab = 20)

and then c = 18k (bc = 18)

space diagonal squared = $d^2 = a^2+b^2+c^2 = 724k^2 + \frac{1}{k^2} \geq 2 \times \sqrt{724} ; (A.M \geq G.M)$

Therefore min. value of d = $\sqrt{2\sqrt{724}} = 7.3358$

Its acually much simpler then what many of you are saying. It is simply that the smallest perimeter for any are of a rectangle is a square. Therefore the sidelengths of the two squares are √20 and √18 . Using the Pythagorean theurum, you get √20² + √18² = C² . Cross out the sqrts and squares to get 20 + 18 = C². 38 = C² C = √38 Now using pythagorus again using the height now that you have the diagonal of the ground segment, you want to use that diagonal with the height. Which is √18² + √38² = C². 18 + 38 = C² --> C = √58 or 7.3358...

Let "a" and "b" be the length of the rectangle which area is 20 Let "a" and "c" be the length of the rectangle which area is 18 ab = 20; ac = 18; b/c = 20/18;

Diagonal D would be: D^2 = a^2 + b^2 + c^2 D^2 = (18/c)^2 + (10c/9)^2 + c^2} ; manipulating to have just one variable Using Minima and Maxima theorem, set dD/dc to 0 you will get: c = 3.47

Substituting c to the Diagonal D equation....

D = 7.335836434

Let $l,w,h$ be the length, width, and height of the rectangular cuboid. This allows us to write:

$lw=20$

$hw=18$

Now, we write $d$ all in terms of our edges, $l, w,$ and $h$ :

$d=\sqrt{l^{2}+w^{2}+h^{2}}$

We can further simplify this equation by recognizing that $l$ and $h$ can be written in terms of $w$ , cutting our variable count from 3 to 1:

$l=\frac{20}{w}$

$h=\frac{18}{w}$

$d=\sqrt{(\frac{20}{w})^{2}+w^{2}+(\frac{18}{w})^{2}}$

Then we expand the two fractions and add the like terms:

$d=\sqrt{w^{2}+\frac{724}{w^{2}}}$

Next, we can find the minimum value of the variable in a quadratic equation by finding when the the first derivative of that equation equals 0. To make this a bit easier to work with, we'll square both sides of the above equation, and then let $f(w)=d^{2}$ for conventions:

$d^{2}=w^{2}+\frac{724}{w^{2}}=f(w)$

Next, we evaluate the first derivative of $f(w)$ , $f'(w)$ (note: I rewrote $\frac{724}{w^{2}}$ as $724w^{-2}$ beforehand as a personal preference; I find it makes differentiating easier):

$f'(w)= 2w-1448w^{-3}$

Now that we have our first derivative, we need to solve for when $f'(w)=0$ . This will happen when $2w = 1448w^{-3}$ . The rest is algebra:

$2w=1448w^{-3}$

$2w^{4}=1448$

$w^{4}=724$

$w=\pm5.18721...$

Since the side of a cuboid cannot be negative, the minimum value of our width, $w$ , is approximately $5.18721$ . This can then be plugged back into our equation for $d$ to get our final answer:

$d=\sqrt{5.18721^{2}+\frac{724}{(5.18721)^{2}}}=7.3358$

It is easy to note that, ac=20; bc=18;

Now, we need to find the diagonal length which is √a^2 + b^2 + c^2 (It can be obtained using pythagoras theorem). Applying AM - GM inequality in terms (a^2 + b^2) and c^2; - We come up with {a^2 + b^2 + c^2 }/2 ≧ √ [ {a^2 + b^2} * c^2 ] ; so, a^2 + b^2 + c^2 ≧ 2 * √ (ac)^2 + (bc)^2 ; Putting values and solving, Least value of a^2 + b^2 + c^2 is 53.814; But d = √a^2 + b^2 + c^2 ; = √53.814 ; = 7.3358 ;

The minimum length of d is 7.33583643 when b=5.1872.
bc=18, c=18/b
ab=20, a=20/b
d^2=a^2+b^2+c^2
d^2=400/b^2+b^2+324/b^2 . Instead of finding derivative and then calculate the value,
I decided to write a small program.
First I ran b from 1 to 9 with step 1
Then - from 4 to 6 with step .1 and couple more times ... ,
This way I narrowed the range for b and visually found the value for b with minimal d.

           for b = 5.185 to 5.189 step .0001   
               b2 = b^2   
               d2 = (400/b2)+b2+(324/b2)   
               d = sqr(d2)   
               print b,d   
           next b

We need to minimize $d=\sqrt{a^2+b^2+c^2}$ , where $a,b,c$ are the 3 sides of the cuboid. We see that the area condition can be reformulated so they're dependent on 1 variable. $a \cdot b=18 \rightarrow b=\frac{18}{a}$ and $a \cdot c=20 \rightarrow c=\frac{20}{a}$ . Now we have a function of 1 variable $f(a) = \sqrt{a^2+(\frac{18}{a})^2 +(\frac{20}{a})^2}$ .

Now we solve the $f'(a)=0$ , and we'll only look at solutions $a \geq 0$ , and we get the solution $a \approx 7.335836434$ .