Answer:
Following are the solution to the given equation:
Step-by-step explanation:
The graph file and correct question are defined in the attachment please find it.
According to the linear programming principle, we predict, that towards the intersections of the constraint points in the viability area, and its optimal solution exists. The sketch shows the points that are (0,16), (3,1), and (6,0).
by putting each point value into the objective function:
Thus, the objective of the function is reduced with a value of 183 at (3,1).
has critical points where the derivative is 0:
The second derivative is
and 
, which indicates a local minimum at 
with a value of 
.
At the endpoints of [-2, 2], we have 
and 
, so that has an absolute minimum of 
and an absolute maximum of 
on [-2, 2].
So we have
The volume, surface area and the ratios of the SA to volume will be as follows:
Volume=πr²h
Area=2πr²+πdh
Ratio of SA to volume=Area/volume
π=3.14
Thus using the above formula:
1.
a]
Radius: 3 inches
Height: 2 inches
Volume=πr²h
volume=π×3²×2=56.52 in³
b]
Area=2πr²+πdh
2×π×3²+π×2×3×2
=56.55+37.68
=94.23 in²
c]
Ratio=area/volume
=94.23/56.52
=1.6672
1.
Radius: 2 inches
Height: 9 inches
a]
V=πr²h
V=3.14*2^2*9
V=113.04 in³
b]
Area=2πr²+πdh
=2*3.14*2^2+3.14*2*2*9
=25.12+113.04
=138.16 in²
c]
Ratio=area/volume
=138.16/113.04
=11/9
3.
Diameter=4 inches
Height= 9 inches
a]
V=πr²h
V=3.14×2²×9
V=113.04
b]
Area=2πr²+πdh
=2*3.14*2^2+3.14*4*9
=25.12+113.04
=138.16 in²
c]
Ratio=area/volume
=138.16/113.04
=11/9
4]
Diameter: 6 inches
Height: 4 inches
a]
Volume=πr²h
=3.14×3²×4
=113.04 in³
b]
Area=2πr²+πdh
=2×3.14×3²+3.14×6×4
=56.52+75.36
=131.88 in²
c] Ratio
131.88/113.04
=7/6
1. For the surface area to volume to be small it means that the area is smaller than the volume, for surface area to volume be larger it means that the surface area is larger than the volume. It is more economical for the surface area to volume to be small because it will mean that small amount of materials make cans with large volume. This means cost of production is cheaper.
2. To evaluate this process let's use one of the dimensions:
Radius: 3 inches
Height: 2 inches:
i. add radius and height:
3+2=5 inches
ii. Multiply radius and height:
3×2=6
iii. Dividing the result from step 1 by the result in step 2:
5/6
iv. Multiply the result from step 3 by 2:
5/6×6
=5
This result does not seem to add up to the result in our earlier ratio. Thus we conclude that Khianna was wrong. This method can't work with 3-D figures.
