Abox (with no top) is to be constructed from a piece of cardboard of sides 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 volume of the box if a = 8 and b = 9.

The volume of the box is h(8-2h)(9-2h). Differentiating this gives dV/dh = 12h^2-68h+72. The derivative is zero when h = (17+sqrt(73))/6 or (17-sqrt(73))/6. The second derivative is positive for the first value, so this gives a minimum volume. The second derivative is negative for the second value, so this gives the maximum volume. So the value of h that maximizes the volume of the box is (17-sqrt(73))/6, which is approximately 1.41.

The answer is 270 cars.you can do this by making a proportional relationship. for instance, 200 is the whole we are using, so on a “number line” 200 would be over 100%. now, since he sells 35% more, add 35% to 100% to get 135% on the other side of the “number line” on the bottom half, leaving the top half empty.cross multiply 200 by 135 then divide by 100 to get your answer. i hope this , if it did, , if you don’t mind, giving me brainiest! 200/100% } 200 • 135 /100 ? /135%