max/min probs: fencing a garden; sliding ladder

[Aibetsu]

1) A 200 ft² rectangular garden is enclosed by a fence divided into two equal parts by another fence parallel to one of the sides.
a.) What dimensions of the outer rectangle will require the smallest total length of fence?
b.) How much fence is needed?

I tried this one out and I got 16 x 12.5, I couldnt find anything else that would be smaller, am I correct?

2) A 17 foot ladder is leaning against a house, whose base starts to slide away. By the time the base is 8ft from the house, the base is moving away from the wall at a rate of 6ft/sec.
a.) How fast is the top of the ladder sliding at that moment?
b.) What rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment?
c.) At what rate is the angle between the ladder and the ground changing at that moment?

I tried it out and for part a) I got 8 ft/sec, for part b) I got -14 ft/sec and for part c) I got -1 rad/sec. Am I correct?

Thank you so much!!

 

[galactus]

Re: I need help with two word problems please!

A 17 foot ladder is leaning against a house, whose base starts to slide away. By the time the base is 8ft from the house, the base is moving away from the wall at a rate of 6ft/sec.
a.) How fast is the top of the ladder sliding at that moment?

You can use Pythagoras, of course.

\(\displaystyle \L\\x^{2}+y^{2}=289\)

\(\displaystyle \L\\x\frac{dx}{dt}+y\frac{dy}{dt}=0\)

\(\displaystyle \L\\\frac{dy}{dt}=-\frac{x}{y}\cdot\frac{dx}{dt}\)

We know from Pythagoras, when the base is 8 feet away:

\(\displaystyle \L\\y=\sqrt{17^{2}-8^{2}}=15\)

Therefore, \(\displaystyle \L\\\frac{dy}{dt}=\frac{-8}{15}(6)=\frac{-16}{5} \;\ ft/sec\)


Now, use this answer along with the product rule and the formula for the area of a triangle to get part b.

b.) What rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment?

\(\displaystyle \L\\A=\frac{1}{2}xy\)

Use the product rule and fill in the given variables.

c.) At what rate is the angle between the ladder and the ground changing at that moment?

You could start with \(\displaystyle \L\\cos({\theta})=\frac{x}{17}\)

You know all the needed variables. Find \(\displaystyle \frac{d{\theta}}{dt}\)

I tried it out and for part a) I got 8 ft/sec, for part b) I got -14 ft/sec and for part c) I got -1 rad/sec. Am I correct?

Thank you so much!!

No, no, and no.

 

[Aibetsu]

thanks so much for your help!

For part b, do you mean that I simply do this?:

1/2((8)(17)) = 68 ??

And I still do not understand part c, could you help me out some more?

Thank you!

 

[galactus]

thanks so much for your help!

For part b, do you mean that I simply do this?:

1/2((8)(17)) = 68 ??

Come on. It's related rates. No, you can not simply do that. That's fine if you just want to find the area. They want to know at what rate it is changing when x=8.
As I said before, use the product rule on the formula for tha area of a triangle.
Differentiate implicitly.

\(\displaystyle \frac{dA}{dt}=1/2\underbrace{(x\cdot\frac{dy}{dt}+y\cdot\frac{dx}{dt})}_{\text{product rule}}\)

And I still do not understand part c, could you help me out some more?

I got you started in the previous post with cos. Differentiate, sub in your known values and solve for \(\displaystyle \frac{d{\theta}}{dt}\)

Here, \(\displaystyle {-}sin({\theta})\frac{d{\theta}}{dt}=\frac{1}{17}\cdot\frac{dx}{dt}\)

Now, can you finish?.

Thank you!