A wagon can hold 44 pounds. A newspaper weighs 22 ounces. How many newspapers can the wagon hold?
2 answers:
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Let p(t) be the number attendees and t be the ticket price measured in units of 10s of rupees. When the price is t = 7 (i.e. Rs70), there were p(7) = 300 people in attendance.
For each unit increase in t (i.e. for each Rs10 increase in price), p(t) is expected to fall by 20, so that
p(t + 1) = p(t) - 20
Solve for p(t). Suppose t ≥ 7. By substitution,
p(t + 1) = (p(t - 1) - 20) - 20 = p(t - 1) - 2×20
p(t + 1) = (p(t - 2) - 20) - 2×20 = p(t - 2) - 3×20
p(t + 1) = (p(t - 3) - 20) - 3×20 = p(t - 3) - 4×20
and so on, down to
p(t + 1) = p(7) - (t - 6)×20 = 420 - 20t
or
p(t) = 420 - 20 (t - 1) = 440 - 20t
With t = price per ticket (Rs/ticket) and p(t) = number of attendees = number of tickets sold, it follows that the income made from ticket sales for some fixed ticket price t would be t×p(t). If one plots p(t) in the coordinate plane, the price that maximizes income and number of attendees is such that the area of a rectangle inscribed by the line p(t) and the coordinate axes is maximized.
Let A(t) be the area of this rectangle, so
A(t) = t p(t) = 440t - 20t²
Without using calculus, complete the square:
A(t) = 440t - 20t² = 2420 - 20 (t - 11)²
This is the equation of a parabola with vertex at (11, 2420), so the optiml ticket price is Rs11, and at this price the drama club can expect an income of Rs2420.
With calculus, differentiate A with respect to t and find the critical points:
A'(t) = 440 - 40t = 0 ⇒ 11 - t = 0 ⇒ t = 11
Differentiate A again and check the sign of the second derivative at this critical point:
A''(t) = -40 ⇒ A''(11) = -40 < 0
which indicates a local maximum at t = 11 of A(11) = 2420.
This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.
Complete Question
Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)
A) 1/3 times
B) 3 times
C) 9 times
D) The same amount of aluminum will be required.
Answer:
B) 3 times
Step-by-step explanation:
From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.
From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included
The curved surface area of a cylinder is calculated as 2πrh
We told that both cylinders have the same height. Hence,
For Dewar A
We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 15 ×h
= 30πhcm²
For Dewar B
We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.
Height = h
The curved surface area = 2πrh
= 2 × π × 5 ×h
= 10πhcm²
When we compare the curved surface Dewar A to the curved surface area of Dewar B
Dewar A : Dewar B
30πhcm² : 10πhcm²
3(10πhcm²) : 10πhcm²
From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.
