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Math  /  Algebra

Question6 of the questions remain unanswered.
A box is to be made out of a 12 cm by 20 cm piece of cardboard. Squares of side length x cmx \mathrm{~cm} will be cut out of each corner, and then the ends and sides will be folded up to (a) Express the volume VV of the box as a function of x\boldsymbol{x}. V=□cm3V=\square \mathrm{cm}^{3} (b) Give the domain of VV in interval notation. (Use the fact that length and volume must be positive.) □\square (c) Find the length LL, width WW, and height HH of the resulting box that maximizes the volume. (Assume that W≤LW \leq L ). L=□cmW=□cmH=□cm\begin{array}{l} L=\square \mathrm{cm} \\ W=\square \mathrm{cm} \\ H=\square \mathrm{cm} \end{array} (d) The maximum volume of the box is □cm3\square \mathrm{cm}^{3}.

Studdy Solution

STEP 1

1. The cardboard is a rectangle with dimensions 12 cm by 20 cm.
2. Squares of side length x x cm are cut from each corner.
3. The box is formed by folding up the sides after cutting.

STEP 2

1. Express the volume V V of the box as a function of x x .
2. Determine the domain of V(x) V(x) .
3. Find the dimensions L L , W W , and H H that maximize the volume.
4. Calculate the maximum volume of the box.

STEP 3

To express the volume V V as a function of x x , consider the dimensions of the box after cutting and folding: - Length of the box: 20−2x 20 - 2x - Width of the box: 12−2x 12 - 2x - Height of the box: x x
The volume V V is given by: V(x)=(20−2x)(12−2x)x V(x) = (20 - 2x)(12 - 2x)x

STEP 4

To determine the domain of V(x) V(x) , consider the constraints: - The length 20−2x 20 - 2x must be positive: 20−2x>0 20 - 2x > 0 implies x<10 x < 10 . - The width 12−2x 12 - 2x must be positive: 12−2x>0 12 - 2x > 0 implies x<6 x < 6 . - The height x x must be positive: x>0 x > 0 .
The domain of V(x) V(x) is 0<x<6 0 < x < 6 .

STEP 5

To find the dimensions L L , W W , and H H that maximize the volume, we need to maximize the function V(x)=(20−2x)(12−2x)x V(x) = (20 - 2x)(12 - 2x)x .
First, expand and simplify the volume function: V(x)=4x3−64x2+240x V(x) = 4x^3 - 64x^2 + 240x
Find the critical points by taking the derivative and setting it to zero: V′(x)=12x2−128x+240 V'(x) = 12x^2 - 128x + 240
Set V′(x)=0 V'(x) = 0 and solve for x x : 12x2−128x+240=0 12x^2 - 128x + 240 = 0
Simplify and solve the quadratic equation: x2−323x+20=0 x^2 - \frac{32}{3}x + 20 = 0
Use the quadratic formula: x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=323±(323)2−4⋅1⋅202⋅1 x = \frac{\frac{32}{3} \pm \sqrt{\left(\frac{32}{3}\right)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1}
Calculate the solutions for x x within the domain 0<x<6 0 < x < 6 .

STEP 6

Calculate L L , W W , and H H using the value of x x that maximizes the volume: - L=20−2x L = 20 - 2x - W=12−2x W = 12 - 2x - H=x H = x
Verify W≤L W \leq L .

STEP 7

Substitute the value of x x back into the volume function to find the maximum volume: Vmax=(20−2x)(12−2x)x V_{\text{max}} = (20 - 2x)(12 - 2x)x
Calculate the maximum volume.
The solutions are: (a) V(x)=4x3−64x2+240x V(x) = 4x^3 - 64x^2 + 240x cm3^3 (b) Domain: (0,6) (0, 6) (c) L=□ L = \square cm, W=□ W = \square cm, H=□ H = \square cm (d) Maximum volume: □\square cm3^3

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