Geometry and Optimization
1. Solve for x and y.
Solution: The sum of the interior angles in a quadrilateral is 360
72+y+65+90=360
y+227=360
y=360-227
y=133
Find x by using the smaller quadrilateral.
y+95+72+ x=360
133+ 95+72+x=360
300+x=360
x=360-300
x=60
x=60, y=133
2. Determine the number of sides in a regular polygon if the sum of the interior angles is 5040
Solution: Recall that the formula for the sum of the interior angles in a regular polygon is n-2× 180
n-2180=5040
180n-360=5040
180n = 5040+360
180n = 5400
180n 180= 5400 180
n=30
Or
n-2180=5040
n-2180180=5040180
n-2=28
n=28+2
n=30
This polygon has 30 sides
3. Golf balls are stacked 3 high in a box. The radius of one ball is approximately 2.1 cm.
a) Draw and label a diagram.
b) What is the minimum volume of the box needed to accommodate the 3 golf balls?
c) What is the volume of the empty space?
Sol: a) r=2.1 cm
d=2r
=2 ×2.1
=4.2
h=3 ×4.2
=12.6 cm
b) Vbox=l w h
=4.2 ×4.2 ×12.6
=222.26
The minimum volume of the box needed to accommodate three GB is approve 222.
26 cm^3
c) Volume of three golf balls
VEmpty spare=VBox-V3GB
=222.26-116.38
=105.88
V3 GB=343r3
=4 πr3
=4π2.13
=37.044
=116.38
4. . A 3 L box of rice is a squarebased prism and is to be made from the minimum amount of cardboard
a) Determine the dimensions of the box that requires the least amount of cardboard material. Round to the nearest tenth of a centimetre. Hint: 1 L = 1000 cm3.
b) How much cardboard material is required to make one box? Round to the nearest cm2. c) If cardboard costs per 0.008 cm2 , what is the price of material for 210 boxes?
Solution: a) V=3L
=3000 cm3
3L=3 ×1000 cm3
=3000 cm3
Vcube=S3
3000=S3
33000=3S3
S=33000
S=14.4
b)S Acube=6S2
=614.42
=1244.16
=1244
Approx 1244 cm2 of cardboard is required to make this box
c) Cost=210 × 1244×0.008
=2089.92
It will cost $2089.92 to make 210 boxes
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