Findng global and extema

[Hockeyman]

1) Find the point on the curve y= SqRt x on the interval [0,3] that is closest to the point (2,0).
So far i have gotten D=(-y^2 + x^2 - 4x + 4)^(1/2)
Now I'm not sure where to go from there.

2) An open box is to be made from a 16 inch by 30 inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. What size should the squares be to obtain a box with the largest possible volume? what is the largest volume?

If someone could just set up the equation for me, i could find the derivative and do the rest.

 

[Deleted member 4993]

1) Find the point on the curve y= SqRt x on the interval [0,3] that is closest to the point (2,0).
So far i have gotten D=(-y^2 + x^2 - 4x + 4)^(1/2)<--Why do you have negative y^2 - please show steps
Now I'm not sure where to go from there.

2) An open box is to be made from a 16 inch by 30 inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. What size should the squares be to obtain a box with the largest possible volume? what is the largest volume?

Setting up the equation is 90% of the work - rest is just routine. Anyway,

start by assuming

sides of the square = h

Now draw a picture. If you cut squares of size (h*h) from the four corners - what does it look like?

Then if if you bend those flaps - what happens?

Tell us with your reply.


If someone could just set up the equation for me, i could find the derivative and do the rest.

 

[Hockeyman]

on the first problem that should be a positive

for the second problem, if you cut out the squares of size h*h from the four corners and bring the flaps up then you have a box. so you removed h*h which is h^2 four times, so that would be 4h^2. Now how would you derive the equation?

 

[Deleted member 4993]

When you bring the flaps up - you'll have rectangular-bottom box with a height of 'h'.

What are the dimensions of the bottom of this box? (after you brought the flaps up)

 

[Deleted member 4993]

on the first problem that should be a positive --> so fix it - then replace y by \(\displaystyle sqrt{x}\)

Now you have D as function of 'x' only.

Now what are you supposed to do to "minimize" the distance.

for the second problem, if you cut out the squares of size h*h from the four corners and bring the flaps up then you have a box. so you removed h*h which is h^2 four times, so that would be 4h^2. Now how would you derive the equation?