Container for cake boxe

Question

Container for cake pieces )
A manafacture supplies bakery treats with closed boxes . A right prism of height h cm. The cross section is a sector of a circle with radius r cm and angle 1 radian the volume of the box is 300 cm cube what dimensions should the manufacturer use to optimize the minimum amount of materials. 

  

 

 

 

 

(How do i go about doing this question the graphical method help)

Answer 1

 

You have two constraints on the box;

The volume, V = 300 cm³

The angle, θ = 1 radian

* Radian: the angle made when we take the radius and wrap it round the circle (arc length).

 

Start by finding the equation for the volume.

 

Area of sector of circle = (1/2)r²θ  

θ must be in radians.

 

The area of the sector, (base) times the height will give us the volume.

 

V = hr²θ/2 = 300 cm³

 

Solve for the height in terms of r. θ = 1 radian

 

h = 600 cm³/r²

 

The question asks how we can minimize the materials used to make the box. This means we need a minimum for the surface area of the box with the constraints.  

 

The surface area of the box is the area of all sides.

The top and bottom of the box is the sector

2* (1/2)r²θ = r²θ = r²

θ = 1 radian

 

The three rectangular sides are the same because the angle, 1 radian tells us that the arc length = radius. So the combined area of those sides will be;

3 * hr

 

So the surface area of the box, S = 3hr + r²

 

Now substitute h = (600 cm³)/r²  to ensure the volume constraint is met.

 

S = (1800 cm³)/r + r²

 

If you are in calculus, you can take the derivative, (ds/dr) set equal to zero and solve to find the minimum.

 

ds/dr = 2r - 1800r^-2 = 0 

2r = 1800/r²

r³ = 900

r = ³√900

 

If this is a minimum the d²s/dr² > 0 is also a minuimum

2 + 3600r³ > 0

 

Put r value into s and solve for the minimum surface area