Question 18. Suppose we have a piece of cardboard that is 40 cm by 25 cm.

a. The length of the box = (40 - 2·x) cm

The height of the box = x cm

The width of the box = (25 - 2·x) cm

b. The formula for the volume of the box as a function of x is 4·x³ - 130·x² + 1000·x

c. The value of x that would maximize the volume of the box is x = 5 cm

d. The largest volume of the box is 2250 cm³

Step-by-step explanation:

a. The given parameters are;

The length of the cardboard = 40 cm

The width of the cardboard = 25 cm

The length of the box = (40 - 2·x) cm

The height of the box = x cm

The width of the box = (25 - 2·x) cm

b. The formula for the volume of the box = The area of the base of the box × Height of the box

The area of the base of the box =  (40 - 2·x) × (25 - 2·x) = 1000 - 80·x - 50·x + 4·x²

∴ The area of the base of the box = 4·x² - 130·x + 1000

The height of the box = x

The volume of the box = (4·x² - 130·x + 1000) × x = 4·x³ - 130·x² + 1000·x

The volume of the box in terms of x, V = 4·x³ - 130·x² + 1000·x

c. At the extremum point, dV/dx = 12·x² - 260·x + 1000 = 0

x = (260 ± √((-260)² - 4 × 12 × 1000))/(2 × 12)

x = (260 ± 140)/(24)

x = 5 or x = 16.

At x = 5, the volume of the box is V = 4×5³ - 130×5² + 1000×5 = 2250

The volume of the box is V = 2250 cm³

At x = 16.67, the volume is 4×16.67³ - 130×16.67² + 1000×16.67 = -925.cm³

Therefore, the value of x that would maximize the volume of the box is x = 5 cm

d. The largest volume of the box is 4×(5 cm)³ - 130×(5 cm)² + 1000×(5 cm) = 2250 cm³.

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.