How to prove that a cube has a bigger volume than a cuboid with the same sum of edge lengths?
Example:
Cube has an edge length of 5 cm. $\ V =5 \times 5 \times 5 = 125 cm^2$
Cuboid has an edge lengths of 3, 5, 7 cm. $\ V =3 \times 5 \times 7 = 105 cm^2$
$3 + 5 + 7 = 5 + 5 + 5 = 15$
$ 125 > 105 $
On
@user635053 Then, Dear Friend, Let me show you a algebraic way. let U be the "same" sum of edge lengths. Let the cube have edges of length x. And for the cuboid let them be x, x+a, and x+b. Since the sum of the edge lengths must be equal, x+x+a+x+b = 3x . Then this simplifies into a = -b. Now, Let V be the volume i.e. V= x(x+a)(x+b) It should be sufficient to show that the volume is greatest when a and b both equal 0 V = x(x+a)+(x+b) =x²+ax(x+b) = x³ + x²b + ax²+ abx = x³ + (b+a)x² + abx, since a= -b, b+a =0. Thus this equals x³+abx. Now since a and b have opposite signs ab is always negative, it equal -a² or -b² to be exact. And now it is evident that the maximum vamue ab can reach is 0. This it reaches when a and b both equal zero i.e. the cuboid is the cube. One more way Let the edges be a, b and c Then let U be their sum I.e. U = a+b+c This we can also write as U = (a+b)+c Now let V be the volume. That is, V = abc = (ab) c. Now let a+b = P. Then it can be shown that ab is macimum when a=b=P/2 [ the proof is here: a+b=P => b=P-a. Thus ab = aP-a². This equation is of a parabola and we know that the vertex of the parabola is the maxima. The vertex by viete's formula (this things are very trivial and the reader can study them on his/her own, so to keep it concise and readable, we need not go into much detail there) is given as -b/2a where b is the coefficient of the linear term (a) and a the coefficient of the quadratic term (a²). P is the coefficient of the linear term and -1 is the coefficient of the quadratic term. Thus the vertex or maxima is at a = P/2. And then solving for b we have b = P-P/2 = P/2. Now abc is maximum when (ab) and c are both maximum. So we just saw that ab is maximum when a and b are both half of their sum. Let them be subsituted with a single variable for they are equal. We can call b0th a or both b; I will call both a, . Then V = a²c. But now U = a+b+c = 2a+c. Thus c = U-2a. Thus, V= a²(U-2a) = a²U-2a³. Now U is constant, so differentiate it with respect to a, we get 2au-6a². We know that a function has slope 0 at its maxima. So 2au-6a² must be 0. I.e. 2au=6a² => u =3a. Now since u also equals 2a+c, c= u/3 and so do a and b. Hence proved that all the edges are equal.
Let the cuboid have sides $a,b,c > 0$ and constrain $a+b+c=K$ for a constant $K$. Now optimize the choice of $a,b,c$ to maximize the volume.
You will get $a=b=c=K/3$, yielding your result.
You have $c = K - a -b$ and $$V(a,b) = ab(K-a-b) = Kab - a^2 b - ab^2,$$ which we seek to maximize. Note that $$ \begin{split} V_a(a,b) = \frac{\partial V}{\partial a} &= Kb - 2ab - b^2 = b(K-2a-b)\\ V_b(a,b) = \frac{\partial V}{\partial b} &= Ka - 2ab - a^2 = a(K-2b-a)\\ \end{split} $$ so solving $V_a(a,b) = 0 = V_b(a,b)$, the first equation yields $b=0$ (which is the minimum solution of zero volume) and $2a+b=K$, whereas $V_b(a,b)=0$ yields $2b+a = K$. Hence you have $$ 2a+b = K\\ a + 2b = K, $$ which has a unique solution at $a=b=K/3$.