Using Minimize of Area,
the dimensions of box is (50cm , 50cm ) that minimize the amount of material used.
We have required box is shape of square base and open top i.e cube without top
Volume of box (V) = 62500 cubic cm
Let us consider, dimensions of the base of the box be (b cm X b cm).
using , Volume = area× height
so, height of the box = {62500 / (b²)} cm.
Total surface area of the top-open box = S
= [(b²) + 4b * {62500 / (b²)}] cm²
= {(b²) + (250000 / b)} cm².
therefore, S = {b² + (250000 /b)}……………(1)
we have the area of box is minimize,
then for extreme values of S, we first calculate first order derivative of S with respect to "b" and put (dS / da) = 0.
Now, (dS / db) = [2b - {250000 / (b²)}]…………(2)
Therefore, for (dS / db) = 0
2a = 250000 / (b²)
(b³) = 125000 = (50)³
=> b = 50 cm.
Height = {62500 / (50^2)} cm = 25 cm.
For S = S min; (d² / db²) S > 0.
Now, (d²/ db²) S = [2 + {500000 / (b³)}] is greater than 0 for any positive value of x.
Therefore, S = S min, when b = 50 cm.
S min = {(50²) + (250000 / 50)} cm²= (2500 + 5000) cm² = 7500 cm².
We know that smallest amount of material will be required to make the box when its total surface area will be minimum.
Therefore, to minimize the amount of materials used, dimensions of the box are ( 50 cm, 50cm).
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