An open box is to be made out of a 6-inch by 14-inch piece of cardboard by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume

The dimensions of the resulting box that has the largest volume is 1.3 inches x 1.3 inches

Step-by-step explanation:

Card board size is L= 14 inches and

W = 6 inches

Let x be the size of equal squares cut from 4 corners and bent into a box whose size is now;

L = 14 − 2x , W = 6 −2x and h = x inches.

Volume of the box is given as;

V = (14 −2x)(6−2x)x

V =(4x² − 40x + 84)x

= 4x³ − 40x² + 84x.

Now, for the maximum value,

dV/dx =0

Thus,

dv/dx = 12x² - 80x + 84 = 0

Using quadratic formula

x = [-(-80) ± √(-80²) - 4(12 x 84)]/(2 x 12)

x = [80 ± √(6400 - 4032)]/24

x = (80 + 48.66)/24 or (80-48.66)/24

x = 5.36 or 1.31

Looking at the two values, 1.31 would be more appropriate because if we use 5.36,we will get a negative value of the width (W).

Thus, x = 1.31 inches

Let us use the Second Derivative Test to verify that V has a local maximum at x = 1.31.

Thus;

d²v/dx² = 24x - 80 = 24(1.31) - 80 = -48.56

This is less than 0 and therefore, the volume of the box is maximized when a 1.31 inch by 1.3 inch square is cut from the corners of the cardboard sheet.

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,