a box with no top is to be constructed from a piece of cardboard with sides of length A and B by cutting out squares of length h from the corners and folding up the sides find the value of h that maximizes the volukme of the box if A=17 and B=27

The value of h that maximizes the volume of the box is h = 3.36

Step-by-step explanation:

First, the volume of a box of length L, width W, and height H is:

V = L*W*H.

in this case, we start with a rectangle of measures AxB.

We can define that the side with length A will be the length, and the other side will be the width.

As we cut squares of length h in the corners, now we will have that the measures of the box are:

Length = A - 2*h

Width = B - 2*h

Height = h.

And we know that A = 17 and B = 27.

So we can replace those values and get:

Length = 17 - 2*h

Width = 27 - 2*h

Height = h

The volume of this box will be:

V = (17 - 2*h)*(27 - 2*h)*h = 459*h -54*h^2 - 34*h^2 + 4*h^3.

Now we want to maximize this.

V(h) = 4*h^3 - 88*h^2 + 459*h

To find the maximum, let's look at the first derivation of V(h)

V'(h) = 3*4*h^2 - 2*88*h + 459.

We need to look for which values of h this is zero.

0 = 12*h^2 - 176*h + 459

The solutions are given by the Bhaskara equation:

Then the two solutions are:

h = (176 + 95.33)/24 = 11.3

h = (176 -95.33)/24 = 3.36

Now, the problem with the first solution is that if we cut two squares of length 11.3 in side A (2*11.3 > 17) , the amount of material cut is larger than A, so that option can be discarded.

Then the only option is h = 3.36

The value of h that maximizes the volume of the box is h = 3.36

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation:

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it