maxima / minima (1 Viewer)

[ozidolroks]

can someone please help me with these questions ?

15. A surfboard is in the shape of a rectangle and semicircle. The perimeter is to be 4m. Find the maximum area of the surfboard correct to 2 places.

16. A half-pipe is to be made from a rectangular piece of meta; of length x cm. The perimeter of the rectangle is 30m.

a) Find the dimensions of the rectangle that will give the maximum surface area.

b) Find the height from the ground up to the top of the half-pipe with this maximum area, correct to 1 decimal place.

17. Find the least surface area , to the nearest cm2, of a closed cylinder that will hold a voluime of 400 cm 3.

thankx

 

[bored of sc]

15. Let the radius of the semi-cirlce be x and the length of rectangle be y.

Perimeter = pi*2x/2 (semi-circle, circumference = pi*diameter = pi*2x) + 2y (lengths of rectangle) + 2x (other side of rectangle which happens to be the diameter of the circle)

So P = pi*x + 2y + 2x = 4m (given to us in the question)

2y = 4-x(pi+2) --- made 2y the subject and factorised x from other two terms
y = [4-x(pi+2)]/2 --- y is now the subject

Now that we have a value for y we can use that expression to eliminate the y and produce one variable.

The area is given be the semicircle + rectangle.

A = pi*x²/2 (area of semicircle) + 2x{[4-x(pi+2)]/2} (area of rectangle, length of y (which is replaced with stuff inside bracket and width 2x)

A = (pi/2)x²+2x(2-xpi/2-x)
= pi*x²/2+4x-x²*pi-2x²
= (pi*x²-2x²*pi)/2 +4x -2x²
= (-x²*pi)/2 +4x -2x²
bit of algebraic manipulation required above

Now differentiate
dA/dx = -x*pi +4 -4x
= 4-x(pi+4)
dA/dx = 0 for stationary points
thus 4-x(pi+4) = 0
x(pi+4) = 4
x = 4/(pi+4)
= 0.5600991535
= 0.56m

d²A/dx² = -pi-4<0 therefore negative concavity thus maximum turning point so the maximum area is being found

Now sub value for x into Area equation
A = [-(0.56)²pi]/2 + 4(0.56) -2(0.56)²
= 1.120198272
= 1.12m²

Wow, that question is amazingly long.

 

[bored of sc]

16. A half-pipe is to be made from a rectangular piece of meta; of length x cm. The perimeter of the rectangle is 30m.

a) Find the dimensions of the rectangle that will give the maximum surface area.

b) Find the height from the ground up to the top of the half-pipe with this maximum area, correct to 1 decimal place.

a) 2x+2y = 30, letting the length be x and width be y.
y = (30-2x)/2 = 15-x
Area of rectangle = length*width = xy = x(15-x) = 15x-x²

A = 15x-x²
dA/dx = 15-2x
dA/dx = 0 for stationary points
15-2x = 0
x = 7.5

d²A/dx² = -2 < 0 thus negative concavity so maximum turning point = maximum area

but y = 15-x
y = 15-7.5
= 7.5

so the dimensions are 7.5m*7.5m

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

So 2*pi*r/2 = y or x whichever tickles your fancy since they are the same value
pi*r = y
but y = x = 7.5
pi*r = 7.5
r = 2.387324146

But the radius will be the height from the ground to the top so there's your answer.

The height from the ground to the top of the half-pipe will be 2.4m (to 1.d.p).

A diagram helps (see the Maths In Focus textbook).

 

[bored of sc]

17) The volume of a cylinder is V = pi*r²*h
Now we are told the volume is 400cm³
So V = pi*r²h = 400cm³
Now make h the subject

pi*r²h = 400
h = 400/(pi*r²)

Surface area of a cylinder is A = 2*pi*r(r+h)
So sub in h as the above expression.

A = 2*pi*r(r+400/(pi*r²)
= 2pi*r²+800/r

dA/dr = 4*pi*r -800/r²

dA/dr = 0 for stationary points

4*pi*r -800/r² = 0
4*pi*r³-800 = 0
r³ = 800/(4*pi)
r = cuberoot(800/[4*pi])
= 3.9929
= 4cm

d²A/dr² = 4pi +1600r^-3 > 0 when r = 4
thus positive concavity = minimum turning point = minimum surface area

Remember area = 2pir²+800/r²
when r = 4 A = 8pi+200 = 301cm² (to nearest cm²)

Yay, all done.

Be sure to show unrounded answers first as I haven't done above.

 

[nuvardex]

wow and the asshole who posted this question ghosted you.