Question

repeat creating a box activity and find the maximum volume with the following cardboard dimensions: 10cm x 12cm = 12cm x 18cm = 25cm x 50cm = v=(20 - 2x)(15 - 2x)x

Explanation:

1. For the cardboard of dimensions \(10\mathrm{cm}\times12\mathrm{cm}\):

• First, establish the volume formula. If we cut out squares of side - length \(x\) from each corner of the cardboard to form a box, the length \(l = 10 - 2x\), the width \(w=12 - 2x\), and the height \(h = x\). So the volume \(V(x)=(10 - 2x)(12 - 2x)x\).
• Expand the function:
• \((10 - 2x)(12 - 2x)x=(120-20x - 24x + 4x^{2})x=(120 - 44x + 4x^{2})x=120x-44x^{2}+4x^{3}\).
• Then, find the derivative of \(V(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
• \(V^\prime(x)=120-88x + 12x^{2}\).
• Set \(V^\prime(x)=0\) to find the critical points:
• \(12x^{2}-88x + 120 = 0\). Divide through by \(4\): \(3x^{2}-22x + 30=0\).
• Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\), here \(a = 3\), \(b=-22\), \(c = 30\).
• \(x=\frac{22\pm\sqrt{(-22)^{2}-4\times3\times30}}{2\times3}=\frac{22\pm\sqrt{484 - 360}}{6}=\frac{22\pm\sqrt{124}}{6}=\frac{22\pm2\sqrt{31}}{6}=\frac{11\pm\sqrt{31}}{3}\).
• \(x_1=\frac{11+\sqrt{31}}{3}\approx\frac{11 + 5.57}{3}\approx5.52\) (rejected since \(10-2x\gt0\) and \(12 - 2x\gt0\), so \(x\lt5\)).
• \(x_2=\frac{11 - \sqrt{31}}{3}\approx\frac{11 - 5.57}{3}\approx1.81\).
• Substitute \(x\approx1.81\) into \(V(x)\):
• \(V(1.81)=(10 - 2\times1.81)(12 - 2\times1.81)\times1.81=(10 - 3.62)(12 - 3.62)\times1.81 = 6.38\times8.38\times1.81\approx97.6\mathrm{cm}^{3}\).

1. For the cardboard of dimensions \(12\mathrm{cm}\times18\mathrm{cm}\):

• The volume formula is \(V(x)=(12 - 2x)(18 - 2x)x\).
• Expand it: \((12 - 2x)(18 - 2x)x=(216-24x - 36x + 4x^{2})x=(216 - 60x + 4x^{2})x=216x-60x^{2}+4x^{3}\).
• Find the derivative: \(V^\prime(x)=216-120x + 12x^{2}\).
• Set \(V^\prime(x)=0\): \(12x^{2}-120x + 216 = 0\). Divide through by \(12\): \(x^{2}-10x + 18=0\).
• Using the quadratic formula \(x=\frac{10\pm\sqrt{100 - 72}}{2}=\frac{10\pm\sqrt{28}}{2}=\frac{10\pm2\sqrt{7}}{2}=5\pm\sqrt{7}\).
• \(x_1 = 5+\sqrt{7}\approx7.65\) (rejected since \(12-2x\gt0\) and \(18 - 2x\gt0\), so \(x\lt6\)).
• \(x_2=5 - \sqrt{7}\approx2.35\).
• Substitute \(x\approx2.35\) into \(V(x)\):
• \(V(2.35)=(12 - 2\times2.35)(18 - 2\times2.35)\times2.35=(12 - 4.7)(18 - 4.7)\times2.35 = 7.3\times13.3\times2.35\approx228.1\mathrm{cm}^{3}\).

1. For the cardboard of dimensions \(25\mathrm{cm}\times50\mathrm{cm}\):

• The volume formula is \(V(x)=(25 - 2x)(50 - 2x)x\).
• Expand it: \((25 - 2x)(50 - 2x)x=(1250-50x - 100x + 4x^{2})x=(1250 - 150x + 4x^{2})x=1250x-150x^{2}+4x^{3}\).
• Find the derivative: \(V^\prime(x)=1250-300x + 12x^{2}\).
• Set \(V^\prime(x)=0\): \(12x^{2}-300x + 1250 = 0\). Divide through by \(2\): \(6x^{2}-150x + 625 = 0\).
• Using the quadratic formula \(x=\frac{150\pm\sqrt{(-150)^{2}-4\times6\times625}}{2\times6}=\frac{150\pm\sqrt{22500 - 15000}}{12}=\frac{150\pm\sqrt{7500}}{12}=\frac{150\pm50\sqrt{3}}{12}=\frac{75\pm25\sqrt{3}}{6}\).
• \(x_1=\frac{75 + 25\sqrt{3}}{6}\approx\frac{75+43.3}{6}\approx19.72\) (rejected since \(25-2x\gt0\) and \(50 - 2x\gt0\), so \(x\lt12.5\)).
• \(x_2=\frac{75 - 25\sqrt{3}}{6}\approx\frac{75 - 43.3}{6}\approx5.28\).
• Substitute \(x\approx5.28\) into \(V(x)\):
• \(V(5.28)=(25 - 2\times5.28)(50 - 2\times5.28)\times5.28=(25 - 10.56)(50 - 10.56)\times5.28 = 14.44\times39.44\times5.28\approx2980.5\mathrm{cm}^{3}\).

Step1: Establish volume functions

For each cardboard size, create a volume function \(V(x)=(l - 2x)(w - 2x)x\) where \(l\) and \(w\) are the length and width of the cardboard.

Step2: Expand volume functions

Use the distributive property to expand the functions into polynomial form.

Step3: Find derivatives

Apply the power - rule to find the first - derivative of each volume function.

Step4: Find critical points

Set the first - derivative equal to zero and solve for \(x\) using the quadratic formula.

Step5: Determine valid critical points

Reject critical points that do not satisfy the domain restrictions (\(l-2x\gt0\) and \(w - 2x\gt0\)).

Step6: Calculate maximum volumes

Substitute the valid critical points into the volume functions to find the maximum volumes.

Answer:

For \(10\mathrm{cm}\times12\mathrm{cm}\) cardboard, the maximum volume is approximately \(97.6\mathrm{cm}^{3}\).
For \(12\mathrm{cm}\times18\mathrm{cm}\) cardboard, the maximum volume is approximately \(228.1\mathrm{cm}^{3}\).
For \(25\mathrm{cm}\times50\mathrm{cm}\) cardboard, the maximum volume is approximately \(2980.5\mathrm{cm}^{3}\).