The follow is an example of a mixture
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To minimize the material usage we have to have the volume requested with the minimum surface area.
The volume is:
And the surface is:
From the first equation we get:
I will use k instead of a number just for the conveince.
We plug this into the second equation and we get:
To find the minimum of this function we have to find the zeros of its first derivative.
Sx will denote the first derivative with respect to x and Sy will denote the first derivative with respect to Sy.
Now let both derivatives go to zero and solve the system (this will give us the so-called critical points).
Now we plug in the first equation into the other and we get:
Now we can calculate y:
And finaly we calculate z:
And finaly let's check our result:
The volume is:
And the surface is:
From the first equation we get:
I will use k instead of a number just for the conveince.
We plug this into the second equation and we get:
To find the minimum of this function we have to find the zeros of its first derivative.
Sx will denote the first derivative with respect to x and Sy will denote the first derivative with respect to Sy.
Now let both derivatives go to zero and solve the system (this will give us the so-called critical points).
Now we plug in the first equation into the other and we get:
Now we can calculate y:
And finaly we calculate z:
And finaly let's check our result: