More Maxima/Minima Problems (1 Viewer)

Jmmalic220 · Nov 4, 2016

How should we work out these questions?
Q 16

A half-pipe is to be made from a rectangular piece of metal of length x meters. The perimeter of the rectangle is 30 meters.

Screen Shot 2016-11-04 at 1.37.53 pm.png

(a) Find the derivatives of the rectangle that will give the maximum surface area, correct to 1 decimal place. [Answer: 7.5]
(b) Find the height from the ground up to the top of the half-pipe with this maximum area, correct to 1 decimal place. [Answer: 2.4]

Q 18

The picture frame shown below has a border of 2 cm at the top and bottom and 3 cm at the sides. If the total area of the border is to be 100 cm², find the maximum area of the frame. [Answer: 160 1/6 cm²)

Screen Shot 2016-11-04 at 1.40.04 pm.png

Q 21

X is a point on the curve y=x²-2x+5. Point Y lies directly below X and is on the curve y = 4x-x². The distance d between X and Y is given by d=2x²-6x+5.

Screen Shot 2016-11-04 at 1.40.10 pm.png

(b) Find the minimum distance between X and Y. [Answer: 1/2 unit]

Q 24

Grant is at point A on one side of a 20 m wide river and needs to get to point B on the other side 80 m along the bank as shown.

Screen Shot 2016-11-04 at 1.40.31 pm.png

Grant swims to any point on the other bank and then runs along the side of the river to point B. If he can swim at 7 km/h and run at 11 km/h, find the distance he swims (x) to minimise the time taken to reach point B. Answer to the nearest metre. [Answer: 26 m]

Q 25

A truck travels 1500 km at an hourly cost given by s² + 9000 cents where s is the average speed of the truck.
The cost for the trip is given by C = 1500(s+(9000/s)).
(b) Find the speed that minimises the cost of the trip. [Answer: 95 km/h]
(c) Find the cost of the trip to the nearest dollar. [Answer: $2846]

cookie_dough · Nov 4, 2016

A=xy, P=2x +2y
P=30= 2(x+y)
15=x+y
15-x=y
SO A=(15-x)x (subbed in y=15-x)
A=15x-x^2
A’=15-2x
Area is a maximum/minimum when A’=0
0=15-2x
2x=15
x=7.5
To check that this is a max:
A’’=-2 As this is a negative value, x=7.5 is a maximum

Advice: find the equations for what you are trying to derive. Do you know how to differentiate? If you do, you use differentiation to find the maximum/minimum values of a coefficient, by differentiating and then making the result = to 0 and solving for the coefficient (e.g. x). Then you differentiate the first differentiate to find the second differentiate. You sub in the coefficient's value into the second differentiate equation to get a value. if the value is positive, the x value is a minimum, if it's negative, the x value is a maximum. In some cases, such as Q21, you will need to sub in x value back into the equation to find the value of d, but it depends on the question.
Hope this helps

cookie_dough · Nov 4, 2016

i just did q18 ceebs the rest, just finished the HSC i need a break

Jmmalic220 · Nov 4, 2016

cookie_dough said:
i just did q18 ceebs the rest, just finished the HSC i need a break

Enjoy your break! Hope you get the top marks!

davidgoes4wce · Nov 5, 2016

$Q16 (a)$

$$ A=xy , where A is the area $ \textcircled{1}$

$$ P=2x+2y , where P is the perimeter $ \textcircled{2}$

$Perimeter=30$

$30=2x+2y$

$y=15-x$

$sub \textcircled{2} into \textcircled{1}$

$A=x(15-x)=15x-x^2$

$\frac{dA}{dx}=15-2x$

$To determine the maximum or minimum , set the first derivative equal to 0.$

$\frac{dA}{dx}=0$

$0=15-2x$

$\frac{d^2A}{dx^2} < 0, $ this negative concavity, so maximum turning point= maximum area. $$

$(b)$

$Now the length/width is equal to the circumference of semi-circle formed.$

$\frac{2 \pi \ r}{2}=y$

$\pi r=y$

$\pi \ r =7.5$

$\therefore $ r=2.4 (1 decimal place) $$

davidgoes4wce · Nov 5, 2016

Jmmalic220 said:
How should we work out these questions?

Q 18

The picture frame shown below has a border of 2 cm at the top and bottom and 3 cm at the sides. If the total area of the border is to be 100 cm², find the maximum area of the frame. [Answer: 160 1/6 cm²)
View attachment 33568

Do you mind if I asked where you got Q18 from?

Jmmalic220 · Nov 6, 2016

This is from the Maths in Focus textbook.

davidgoes4wce said:
Do you mind if I asked where you got Q18 from?

davidgoes4wce · Nov 6, 2016

Jmmalic220 said:
This is from the Maths in Focus textbook.

It's a good book. Great for foundation knowledge.

He-Mann · Nov 6, 2016

davidgoes4wce said:
It's a good book. Great for foundation knowledge.

Good is a vague term, please define it. Otherwise, you may guide some mindless students in the wrong direction.

Mahan1 · Nov 27, 2016

Q24:
Total distance travelled is:

$x+80-\sqrt{x^{2}-400}$

$\textrm{time elapsed} \ \ \ \ \frac{x}{7000} + 80 - \frac{\sqrt{x^{2}-400}}{11000} \textrm{note the speeds are in km/h and the distances are in meters}$

$\textrm{differentiate it and equate it to zero} \ \ \frac{1}{7000} - \frac{x}{11000 \sqrt{x^{2}-400}} = 0$

$(\frac{11}{7})^{2} = \frac{x^{2}}{x^{2}-400}$

$(121-49)x^{2} = (11*20)^{2} \ \ \ \ \ \textrm{(1)}$

$x= \frac{220}{6 \sqrt{2}} = \frac{55\sqrt{2}}{3} = 26m \ \ (x \textrm{ is a distance so we only take the positive case)}$

$\textrm{since (1) is a concave up parabola then the stationary point is minimum}$

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