candy box is made from a piece of cardboard that measures 45 by 24 inches. Squares of equal size will be cut out of each comer. The sides will then be folded up to form a rectangular box. What size square should be cut from each corner to obtain maximum volume? inches should be cut away from each corner to obtain the maximum volume. A square with a side of length (Round to the nearest hundredth as needed.)

Answer:

Each square should have 5 inches of side and area = 25 square inches.

Step-by-step explanation:

Candy box is made that measures 45 by 24 inches.

Let the squares of equal size x inches has been cut out of each corner.

The sides will then be folded up to form a rectangular box.

Now we have to find the size of square that should be cut from each corner to obtain maximum volume of the box.

Now the box is with length = (45 - 2x) inches

and width = (24 - 2x) inches

and height = x inches

Volume of the candy box = Length × width × height

V = (45 - 2x)(24 - 2x)(x)

V = x(1080 - 48x -90x + 4x²)

  = x(1080 - 138x + 4x²)

  = 4x³ - 138x² + 1080x

Now we will find the derivative of volume and equate it to zero.

12(x² - 23x + 90) = 0

x² - 23x + 90 = 0

x² - 18x - 5x + 90 = 0

x(x - 18) - 5(x - 18) = 0

(x - 5)(x - 18)=0

x = 5, 18

Now for x = 18 Width of the box will be = (24 - 2×18) = 24 - 36 = -12

Which is not possible.

Therefore, x = 5 will be the possible value.

Therefore, square having area 25 square inches should be cut out from each corner to get the maximum volume of candy box.

Final answer:

The size of the square that should be cut away from each corner to obtain the maximum volume for a box made from a cardboard measuring 45 by 24 inches is 3 inches.

Explanation:

To find the size of the square that should be cut from each corner to obtain the maximum volume, we should first make an equation for the volume of the box. If x is the length of the side of the square, then the dimensions of the box are (45-2x) by (24-2x) by x, thus the volume of the box V is (45-2x)(24-2x)x.

By using calculus, we can find the derivative of this function, set it to zero and solve, this will give the critical points where the maximum and minimum volumes will be.

The derivative is found to be -4x^2 + 138x - 1080. Setting this to zero and solving, we find that x = 3 and x = 90 are the critical points for the maximum and minimum volumes. Since we cannot cut corners more than 24 inches (this would make the width negative), x = 3 inches is the only feasible solution.

So, 3 inches should be cut away from each corner to obtain the maximum volume.

Learn more about Optimization here:

brainly.com/question/37742146

#SPJ3