Answer:
The dimensions are, base ![b=\sqrt[3]{200}](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5D%7B200%7D) , depth
, depth ![d=\sqrt[3]{200}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B3%5D%7B200%7D) and height
 and height ![h=\sqrt[3]{200}](https://tex.z-dn.net/?f=h%3D%5Csqrt%5B3%5D%7B200%7D) .
.
Step-by-step explanation:
First we have to understand the problem, we have a box of unknown dimensions (base  , depth
, depth  and height
 and height  ), and we want to optimize the used material in the box. We know the volume
), and we want to optimize the used material in the box. We know the volume  we want, how we want to optimize the card used in the box we need to minimize the Area
 we want, how we want to optimize the card used in the box we need to minimize the Area  of the box.
 of the box.
The equations are then, for Volume

For Area

From the Volume equation we clear the variable  to get,
 to get,

And we replace this value into the Area equation to get,


So, we have our function  , which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting
, which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting


After solving the system of equations, we get that the optimum point value is ![d=\sqrt[3]{200}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B3%5D%7B200%7D) and
 and  ![h=\sqrt[3]{200}](https://tex.z-dn.net/?f=h%3D%5Csqrt%5B3%5D%7B200%7D) , replacing this values into the equation of variable
, replacing this values into the equation of variable  we get
 we get ![b=\sqrt[3]{200}](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B3%5D%7B200%7D) .
.
Now, we have to check with the hessian matrix if the value is a minimum,
The hessian matrix is defined as,
![H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]](https://tex.z-dn.net/?f=H%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cfrac%7B%5Cpartial%5E2%20A%7D%7B%5Cpartial%20d%5E2%7D%20%26%5Cfrac%7B%5Cpartial%5E2%20A%7D%7B%5Cpartial%20d%20%5Cpartial%20h%7D%5C%5C%5Cfrac%7B%5Cpartial%5E2%20A%7D%7B%5Cpartial%20h%20%5Cpartial%20d%7D%26%5Cfrac%7B%5Cpartial%5E2%20A%7D%7B%5Cpartial%20p%5E2%7D%5Cend%7Barray%7D%5Cright%5D)
we know that,



Then, our matrix is
![H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]](https://tex.z-dn.net/?f=H%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
Now, we found the eigenvalues of the matrix as follow
![det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0](https://tex.z-dn.net/?f=det%28H-%5Clambda%20I%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-%5Clambda%262%5C%5C2%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%3D%284-%5Clambda%29%5E2-4%3D0)
Solving for , we get that the eigenvalues are:
, we get that the eigenvalues are:   and
 and  , how both are positive the Hessian matrix is positive definite which means that the function
, how both are positive the Hessian matrix is positive definite which means that the function is minimum at that point.
 is minimum at that point.