# Thread: Two Optimization Problems ..

1. ## Two Optimization Problems ..

hi ..
please solve these two problems ..

1) A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum voulme when h=1/3 * H (H over 3).

2) Find the rectangle of maximum area that can be inscribed in a right triangle with legs of length 3 and 4 if the sides of the rectangle are parallel to the legs of the triangle (two sides are basically touching the legs of the triangle).

plz If you can, write compeletly soultion ..
all what I need is an ideal solution ..
do not explain any steps ..

I am waiting..

2. Originally Posted by TWiX
hi ..
please solve these two problems ..

1) A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum voulme when h=1/3 * H (H over 3).

2) Find the rectangle of maximum area that can be inscribed in a right triangle with legs of length 3 and 4 if the sides of the rectangle are parallel to the legs of the triangle (two sides are basically touching the legs of the triangle).

plz If you can, write compeletly soultion ..
all what I need is an ideal solution ..
do not explain any steps ..

I am waiting..
1. maximize V= 1/3pir^2

find the critical max

V(h)=pirR^2/H^2(H^2-4Hh+3h^2)=piR^2/3H^2(H-3h)(H-h)

finish up by checking the endpoints

3. Originally Posted by TWiX
hi ..
please solve these two problems ..

1) A cone with height h is inscribed in a larger cone with height H so that its vertex is at the center of the base of the larger cone. Show that the inner cone has maximum voulme when h=1/3 * H (H over 3).

2) Find the rectangle of maximum area that can be inscribed in a right triangle with legs of length 3 and 4 if the sides of the rectangle are parallel to the legs of the triangle (two sides are basically touching the legs of the triangle).

plz If you can, write compeletly soultion ..
all what I need is an ideal solution ..
do not explain any steps ..

I am waiting..
hypotenuse is y=3x/4

one corner must be on y(x)=3x/4

side1=L=4-x
side2=H=y

therefore A=L*H=(4-x)*y=(4-x)*3x/4