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

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph