# Sliding ladder leaning against wall, and a triangle of maximum area

punjabi_monster
Hello,
Below are two questions i am unsure of. :uhh: Can someone please verify my answers and tell me if i am doing something wrong. :yuck: thank-you for your help.

1. A rigid beam 30 m long is leaning against a vertical wall. If the bottom of the beam is pulled horizontally away from the wall at 3 m/s, how fast is the angle between the beam and the ground changing when the bottom of the beam is 18 m from the wall?

sinθ = y/30

cosθ = 18/30
θ= 53.13

A= bh/2
A= [(x)(30sinθ)] / 2
A'= [(30sinθ) (dx/dt) + (x)(30cosθ) (dθ/dt)] / 2
0= [30sin(53.13) (3) + (18)(30cos53.13) (dθ/dt)] / 2
dθ/dt= -6.67 m/s

2.Triangle ABC is inscribed in a semicircle with diameter BC=12 cm. Find the value of angle B that produces the triangle of maximum area.
Hint: An angle inscribed in a semi-circle is a right triangle.

A=bh/2
A=(12cosB)(12sinB)
=72(2cosBsinB)
=72(sin2B)
A'=72(cos2B)
0=72cos2B
cos2B=0
2B=90°
B=45°

i got sin2B using the trig double angle forumla sin2x=2sinxcosx

## Answers and Replies

Homework Helper
There's a very elegant approach for the second problem. No calculus required (!) :

$$A_{\mbox{triangle}} =\frac{AB\cdot AC}{2}$$ (1)

Constraint, following Pythagora's theorem

$$AB^{2}+AC^{2}=12^{2}$$ (2)

(2) can be written

$$\left(AB-AC\right)^{2} =12^{2}-2 AB\cdot AC$$ (2')

U impose that the product (which by (1)is ~Area) be maximum, therefore the RHS would have to be minimum. Since the LHS can be no less than 0, one gets that it has to be 0.

$$AB=AC \Rightarrow \hat{CBA} =45 \mbox{deg}$$

Daniel.

punjabi_monster
thanks...but is the first question right?

Homework Helper
The reasoning is okay. The area of the triangle is indeed constant. I didn't check the numbers. I hope they're okay.

Daniel.

Gold Member
MHB
punjabi_monster said:
=72(sin2B)
A'=72(cos2B)
0=72cos2B

You forgot the chain rule here, but in the end it didn't affect your answer.

$$A = 72\sin{(2B)}$$
$$A' = 72\cos{(2B)}*2$$