Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
1. Let side of square base of the box = x cm
and height of the box = y cm.
Volume = x²y
4000 = x²y
y = 4000/x² ....(1)
The material used for the box Surface Area = Areaof base + 4 times area of sides
Surface Area, A= x² + 4xy
Plug in value from (1) to get
A = x² + 4x(4000/x²)
A = x² + 16000/x
To find optimal value, find derivative and equate to 0
A' = 2x - 16000/x²
0 = 2x - 16000/x²
16000/x² = 2x
x³ = 8000
x = 20 cm
y = 4000/(20)² = 10
Dimension of the box is 20 cm x 20 cm x 10 cm
2. Let width of the base of the box = x meter
then length of the base = 2x meter
and height of the box = y meter
Volume of the box = x(2x)y = 2x²y
10 = 2x²y
y = 5/x² ....(1)
Cost of box material, C = cost of base + cost of sides
C = 20(2x²) + 12(2xy+4xy)
C = 40x² + 72xy
C = 40x² + 72x(5/x²) .........from (1)
C = 40x² + 360/x
C' = 80x - 360/x²
0 = 80x - 360/x²
360/x² = 80x
x³ = 4.5
x = 1.65 m
y = 5/(1.65)² = 1.84 m
Cost, C = 40x² + 360/x = 40(1.65)² + 360/1.65 = $327.08
We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.