a 160-inch strip of metal 20 inches wide is to be made into a small open trough by bending up two sides on the long side , at right angles to the base. the sides will be the same height , x. if the tgrough is to have a maximum volume, how many inches should be turned up on each side? show your work

V = (20 -2x ) * x * (160) , for 0 <= x <=10
V(x) = -320*x^2+3200*x
solve dV/dx = 0
-640*x+3200= 0
3200 = 640x
x = 3200 / 640 = 5
plug in x=10 into V '' (x) = -640
V ''(5) = -640 < 0
The endpoints x= 0, x= 10 produce zero volume.
V(5) = 8000
Therefore x=5 makes a maximum volume on the interval [0,10]
is this right?

your solution and conclusion are correct

Thank you

they said "the longer side" so u should have used 160-2x instead of 20-2x

Yes, your steps and reasoning are correct.

To find the maximum volume of the trough, you correctly set up the volume equation as V = (20 - 2x) * x * 160, where x represents the height of the sides.

Then you derived the derivative of the volume function with respect to x, which gave you dV/dx = -640x + 3200. Setting this derivative equal to zero, you solved for x and found that x = 5.

To determine if this value of x indeed gives a maximum volume, you calculated the second derivative d^2V/dx^2 and evaluated it at x = 5. Since the second derivative was negative, V ''(5) = -640 < 0, you concluded that x = 5 indeed gives a maximum volume.

Finally, you calculated the volume V(5) and obtained V(5) = 8000. This confirms that with both sides bent up by 5 inches, the trough will have a maximum volume.

Well done!

Someone? Be honest, you took that off of peer answer, didn't you?