Question

You are planning to make an open rectangular box from a 38​-in.-by-75​-in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of the box of largest volume you can make this​ way, and what is its​ volume?

Answer 1

Answer:

V = 10384 in³, L = 59 in, w = 22 in

Step-by-step explanation:

Let:

h = the side of the square;

L = the length of the box;

w = the width of the box

V = the box volume.

L = 75 - 2h

w = 38 - 2h

V = h*L*w

V = h*(75 - 2h)*(38 - 2h)

V = h*(2850 - 150h - 76h + 4h²)

V = 4h³ - 226h² + 2850h

The maximum point (the largest volume) is where the derivate is equal to 0

V' = 12h² - 452h + 2850

12h² - 452h + 2850 = 0

Using Bhaskara's equation:

Δ =(- 452)² - 4*12*2850

Δ = 204304 - 136800

Δ = 67504

h = (-(-452) ± √67504)/2*12

h = (452± 259.8)/24

h1 = (452 + 259.8)/24 = 29.66 in

h2 = (452 - 259.8)/24 = 8 in

Let's calcule the V for these both h:

V1 = 4*(29.66)³ - 226*(29.66)² + 2850*29.66

V1 = -9915.3 in³

V2 = 4*(8)³ - 226*(8)² + 2850*8

V2 = 10384 in³

V must be positive, so h = 8 in, V = 10384 in³, L = 59 in, w = 22 in.