Answer:
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.[tex]\bar 6[/tex]
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.[tex]\overline {925}[/tex]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.
