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.
The dimensions of the box that produces the largest volume are;
length = 59 inches
width = 22 inches
height = 8 inches
Volume = 10384 in³
Let h be the height of the box which also serves as the side length of the congruent squares cut out.
Thus;
Length of box; l = 75 - 2h
Width of box; w = 38 - 2h
We know that formula for volume of a box is;
Volume of box; V = lwh
Thus;
V = h(75 - 2h)(38 - 2h)
Expanding this gives;
V = h(2850 - 76h - 150h + 4h²)
V = 4h³ - 226h² + 2850h ---- eq(1)
Differentiating this volume equation gives;
V' = 12h² - 452h + 2850
To get the height that will maximize the volume, we will set this to zero to get;
Setting the equation to zero, we will have
12h² - 452h + 2850 = 0
Using online Quadratic equation solver, we have;
h ≈ 8 in
Now, putting 8 for h in eq 1 gives us;
V = 4(8³) - 226(8²) + 2850(8)
V = 2048 - 14464 + 22800
V = 10384 in³
Thus;
l = 75 - 2h
⇒ l = 75 - 2(8)
l = 59 in
w = 38 - 2h
⇒ w = 38 - 2(8)
w = 22 in
In conclusion, the dimensions of the box that produces the largest volume are; length = 59 inches; width = 22 inches ; height = 8 inches with Volume = 10384 in³
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