maxima / minima (1 Viewer)

[ozidolroks]

can some please help me with these questions ?

18. A picture frame has a border of 2cm at the top and bottom and 3cm at the sides. If the total area of the border is to be 100cm ^2, find the maximum area of the frame.

19. A 3m piece of wire is cut into two pieces and bent around to form a square and a circle. Find the size of the two lengths, correct to 2 decimal places, that will make the total area of the square and circle a minimum.

20. Two cars are travelling along roads that intersect at right angles to one another. One starts 200 km away and travels towards the intersection at 80km/ hour, while the other travels at 120 km away and travels towards the intersection at 60km/ hour.

Show that their distance apart after t hours is given by

d^2=10 000t^2- 46 400t + 54 400 and hence find their minimum distance apart.

Thanks

 

[bored of sc]

Yes. In due time. Draw diagrams. They make it easier to see how to approach the question. It's about making a substitution to eliminate a variable, differentiating, checking derivates and subbing back into equations.

 

[addikaye03]

can some please help me with these questions ?
20. Two cars are travelling along roads that intersect at right angles to one another. One starts 200 km away and travels towards the intersection at 80km/ hour, while the other travels at 120 km away and travels towards the intersection at 60km/ hour.

Show that their distance apart after t hours is given by

d^2=10 000t^2- 46 400t + 54 400 and hence find their minimum distance apart.

Thanks

The first two just require diagrams (as stated) and careful thought.

Q3)D1=200-80t
D2=120-60t

we know have 2 expressions for the distance and both have the same variable. Also, since they are travelling perpendicular to eachother, u can think of that as one travelling along the x-axis and the other along the y-axis ( if u know what i mean). So we can use a coordinate geom. formula

d=rt(x2-x1)^2+(y2-y1)^2

d^2=(x2-x1)^2+(y2-y1)^2
d^2=(200-80t)^2+(120-60t)2...by expanding both terms and collecting like terms.

d^2=10 000t^2- 46 400t + 54 400 #

d(d^2)/dt=20000t-46400... S.P occurs when d/dt=0

20000t=46400
t=2.32

Check that theirs a change in gradient.
Since at t=2.32, d/dt=0 when t=2, d/dt>0 and when t=3, d/dt<0 therefore Min T.P.

N.B It asks for there distance, so u resub into the original equation.
d=24km