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Math  /  Calculus

QuestionGiven a W=10W=10 inch by L=13L=13 inch piece of paper, we will cut out squares (size xx by xx ) from each corner and fold to create an (open top) box. Our goal is to find the size of the cut out square (x)(x), that maximizes the volume of the box.
The length of the box, as a function of τ\tau, is l=13−2x∨σ6l=13-2 x \vee \sigma^{6}
The width of the box, as a function of xx, is w=10−2x∨06w=10-2 x \vee 0^{6}
The volume of the box, as a function of xx, is V=13⋅10⋅x×06x(13−2x)(10−2x)V=13 \cdot 10 \cdot x \times 0^{6} x(13-2 x)(10-2 x) which, after distributing, simplifies to V=130x−46x2+4x3o6V=130 x-46 x^{2}+4 x^{3} \quad o^{6} To determine the value of xx that corresponds to a maximum volume, we need to find V′V^{\prime}. V′=130−92x+12x2V^{\prime}=130-92 x+12 x^{2}
The x\boldsymbol{x} that corresponds to a maximum volume is x=1.8712×0\boldsymbol{x}=1.8712 \times 0 ob 1.8683623129081 inches and the maximum volume is V=V= □\square cubic inches

Studdy Solution

STEP 1

1. We have a rectangular piece of paper with dimensions W=10 W = 10 inches and L=13 L = 13 inches.
2. We cut out squares of size x×x x \times x from each corner and fold the sides to form an open-top box.
3. We need to find the value of x x that maximizes the volume of the box.
4. The volume of the box is given by the function V(x)=(13−2x)(10−2x)x V(x) = (13 - 2x)(10 - 2x)x .

STEP 2

1. Define the volume function of the box.
2. Differentiate the volume function to find critical points.
3. Solve for the critical points.
4. Determine which critical point maximizes the volume.
5. Calculate the maximum volume.

STEP 3

The volume of the box is given by the function:
V(x)=(13−2x)(10−2x)x V(x) = (13 - 2x)(10 - 2x)x
This represents the product of the length, width, and height of the box.

STEP 4

Differentiate the volume function V(x)=(13−2x)(10−2x)x V(x) = (13 - 2x)(10 - 2x)x with respect to x x to find the critical points.
The derivative is:
V′(x)=130−92x+12x2 V'(x) = 130 - 92x + 12x^2

STEP 5

Solve the equation V′(x)=0 V'(x) = 0 to find the critical points:
130−92x+12x2=0 130 - 92x + 12x^2 = 0
Use the quadratic formula:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=12 a = 12 , b=−92 b = -92 , and c=130 c = 130 .

STEP 6

Calculate the discriminant:
b2−4ac=(−92)2−4⋅12⋅130=8464−6240=2224 b^2 - 4ac = (-92)^2 - 4 \cdot 12 \cdot 130 = 8464 - 6240 = 2224
Now, solve for x x :
x=92±222424 x = \frac{92 \pm \sqrt{2224}}{24}
Calculate the roots:
x1=92+222424≈1.868 x_1 = \frac{92 + \sqrt{2224}}{24} \approx 1.868
x2=92−222424≈5.782 x_2 = \frac{92 - \sqrt{2224}}{24} \approx 5.782

STEP 7

Evaluate the volume function at the critical points and endpoints to determine the maximum volume.
The endpoints are x=0 x = 0 and x=5 x = 5 (since x x must be less than half the smaller dimension).
Calculate V(x) V(x) at x1 x_1 , x2 x_2 , and the endpoints:
V(1.868)≈130(1.868)−46(1.868)2+4(1.868)3 V(1.868) \approx 130(1.868) - 46(1.868)^2 + 4(1.868)^3
V(5.782) V(5.782) is not valid since it exceeds the possible x x .

STEP 8

Calculate V(1.868) V(1.868) :
V(1.868)≈130(1.868)−46(1.868)2+4(1.868)3≈97.26 cubic inches V(1.868) \approx 130(1.868) - 46(1.868)^2 + 4(1.868)^3 \approx 97.26 \text{ cubic inches}

STEP 9

The value of x x that maximizes the volume is approximately x=1.868 x = 1.868 inches, and the maximum volume is approximately V=97.26 V = 97.26 cubic inches.
The size of the cut-out square that maximizes the volume is approximately:
1.868 inches \boxed{1.868} \text{ inches}
The maximum volume is:
97.26 cubic inches \boxed{97.26} \text{ cubic inches}

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