If 30,000 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

We are given the material in cm2 units which is also the unit for area, so we divide the value by the number of sides to get the area of each of the 5 sides

30000/5 = 6000cm2

So finding the volume of a cube is Length x beadth x height, which have the same values, because all sides in a cube are equal

Area = 6000cm2

To find the Length( same as height) we find the root of the Area

Root6000 = 77.46

Length = height = 77.46cm

Volume = Length x Breadth x height = Area x height = 6000 x 77.46 = 464760m3

tifanihayyu tifanihayyu · Answer 2 · 2020-01-16T20:54:02+01:00

The largest possible volume of the box is [tex]500.000 cm^3[/tex]

Explanation:

I f 30,000 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

A Box is a polyhedron with all its faces in rectangular shape. The surface area of the box is

[tex](area of base) + 4* (area of one side) = s[/tex]

[tex]30,000 cm^2 = s^2 + 4hs\\[/tex]

Then we can solve the equation of h in terms of s

[tex]4hs = 30,000- s^2 \\h= \frac{30,000- s^2}{4s}[/tex]

By plugging this into the expression for V , we get

[tex]V = \frac{30,000-s^2}{4s} *s^2\\V= 7500s- \frac{1}{4} s^2[/tex]

The volume cannot be negative, therefore V>0 or V=0

The derivative is [tex]\frac{dV}{ds} = 7500-\frac{3}{4} s^2[/tex]

Then find the critical numbers by setting it to zero:

[tex]7500-\frac{3}{4} s^2 = 0\\ s^2=10.000\\s =+- 100[/tex]

The closed interval method is a way to solve a problem within a specific interval of a function. Now we can use the “closed-interval method” to determine the absolute maximum of V

[tex]V (0) = 0\\V (100) = 7500* 100 -\frac{1}{4} * 100^3\\V (100) = 750000 - 250000\\V (100) = 500000[/tex]

Therefore the largest possible volume of the box is [tex]500.000 cm^3[/tex]

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