All boxes with a square base, an open top, and a volume of 220 ft cubed have a surface area given by S(x)equalsx squared plus StartFraction 880 Over x EndFraction , where x is the length of the sides of the base. Find the absolute minimum of the surface area function on the interval (0,infinity). What are the dimensions of the box with minimum surface area?

The absolute minimum of the surface area

At the minimum surface area,

Step-by-step explanation:

Volume of the box =220 cubic feet.

To find the absolute minimum of the surface area function on the interval , we take the derivative of S(x) and solve for its critical points.

Take the cube root of both sides

Therefore, the absolute minimum of the surface area function on the interval , is:

Since the volume of the box =220 cubic feet

The dimensions of the box with the minimum surface area are base length of 7.61 feet and height of 3.8 feet.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10