Calculus Problems again!

[Stokes]

I have two story problems that I completely struggle with and have no idea what to do.. Would anyone be able to offer help on the steps taken to complete this? These are review questions and if I could understand the process in these I would be better for the final coming up tuesday.

A triangle has legs on the positive x- and y- axes aand its hypotenuse passes through the point (4,1). which such triangle has the smallest area.

&

An oil spill has a shape of a circle and its growing at a rate of 100 m^2 per second. at what rate is the radius increasing when its 500m ?

 

[StevenYoo]

ok, i don;t remember how to do the first one.

for the 2nd one:

according the info given, dA/dt (rate of change of the area over time) = 100
the formula for the area of a circle is : A = pi * r^2

so if you differentiate with respect to time (t): dA/dt = 2 * pi * r * dr/dt

the problem says what is the rate of change of the radius (dr/dt) the instant the radius is 500.

so, plugging in, we get:

100 = 2 * pi * 500 * dr/dt

if you solve for dr/dt, you get :

1/10*pi

 

[fatpat268]

the first one is an optimization problem...

I can't recall how to do it off the top of my head without looking at my book though... it's been a while

 

[ngvepforever2]

The first one It's an optimization problem and a popular one at it too. Found the solution in google with not too much work. The solution below was posted by Ernesto David from PointAsk.com forums (so the credit is to him), the only difference with your question is that the hypothenuse is in point (2,1) instead of (4,1)

Draw the figure on paper. It is a right triangle in the 1st quadrant, whose vertices are
O(0,0)
A(0,y)
B(x,0)
In the hypotenuse AB, is point P(2,1).

Need minimum area of right triangle AOP.
We do it by Diffrential Calculus. Set to zero the 1st derivatiove of the area.

K = area of triangle AOP = (1/2)(x)(y) = (1/2)(xy) ---(i)

K = (area of triangle APO) +(area ot triangle OPB)
K = (1/2)(y)(2) +(1/2)(x)(1)
K = (1/2)(2y +x) ----(ii)

K = K
(1/2)(xy) = (1/2)(2y +x)
xy = 2y +x
xy -2y = x
y(x-2) = x
y = x/(x-2)
Substitute that in (i),
K = (1/2)(x)(x/(x-2))
K = (1/2)[(x^2)/(x-2)] ----(iii)

Differentiate (iii) with respect to x.
dK/dx = (1/2)[(x-2)*(2x) -(x^2)*(1)/[(x-2)^2]
dK/dx = (1/2)[x^2 -4x]/[(x-2)^2] ----(iv)

Set that to zero,
Multiply both sides by 2(x-2)^2,
0 = x^2 -4x
So,
x = 0 or 4 when dK/dx = 0.

When x=0, there is no triangle AOP. No K.

When x=4,
Plug that into (iii),
K = (1/2)[(x^2)/(x-2)] ----(iii)
K = (1/2)[(4^2)/(4-2)] = (1/2)[16/2] = 4 sq. units.

Is K = 4 the minimum K?

We check the slope of the graph of K before and after x=4.
This slope is also dK/dx.
Slope = dK/dx = (1/2)[x^2 -4x]/[(x-2)^2] ----(iv)

Say, when x=3,
slope = (1/2)[3^2 -4*3]/[(3-2)^2] = (1/2)[-3]/[1] = -3/2 ---negative.

Say, when x=5,
slope = (1/2)[5^2 -4*5]/[(5-2)^2] = (1/2)[5]/[1] = 5/2 ---positive.

That means the slope of K goes from negative to positive as it passes x=4.
That means at x=4, K is lowest point, or K is minimum.

Therefore, the lowest area for right triangle AOP is 4 sq. units. ---answer.

Regards

ng

 

[tatteredpotato]

Heres my shot at 1:

Since its an optimization, you need an equation and a constraint. From what you have told me your equation is .5xy, because y is your height (y-axis), and x is your width(x-axis). So your figure is a right triangle. So we have:

f(x)=.5xy

Constraint = (y/1)=(x/(x-4)) Since the point on the hypotenuse allows us to make a similar right triangle.
That solved for y is
y=x/(x-4)

Now we combine:

f(x)=.5x(x/(x-4))

Next you take the derivative:
dy/dx= (2x(2x-8)-2x^2)/(2x-8)^2

Now you need to find the zeros of your derivative and determine your min/max points

You get y=0 at x=8, and x=0

Thats as far as i go, all you need to do is set up a sign chart and determine which is a min point on the graph (assuming that is that i haven't made any mistakes yet, which is likely)

EDIT: too slow lol