a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring…

a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 43 ft by 23 ft. the resulting piece of cardboard is then folded into a box without a lid. find the volume of the largest box that can be formed in this way. b. suppose that in part (a) the original piece of cardboard is a square with sides of length s. find the volume of the largest box that can be formed in this way. a. the maximum volume of the box is approximately □ ft³. (round to the nearest hundredth as needed.)

a. squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 43 ft by 23 ft. the resulting piece of cardboard is then folded into a box without a lid. find the volume of the largest box that can be formed in this way. b. suppose that in part (a) the original piece of cardboard is a square with sides of length s. find the volume of the largest box that can be formed in this way. a. the maximum volume of the box is approximately □ ft³. (round to the nearest hundredth as needed.)

Answer

Answer:

a. Approximately 2028.07 $\text{ft}^3$ b. $\frac{2s^{3}}{27}$

Explanation:

Step1: Find volume formula for part a

The length of the box is $l = 43 - 2x$, the width is $w=23 - 2x$ and the height is $h = x$. So the volume $V(x)=(43 - 2x)(23 - 2x)x=4x^{3}-132x^{2}+989x$.

Step2: Differentiate the volume function

$V'(x)=12x^{2}-264x + 989$.

Step3: Set derivative equal to zero

Using the quadratic - formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here $a = 12$, $b=-264$, $c = 989$. $x=\frac{264\pm\sqrt{(-264)^{2}-4\times12\times989}}{2\times12}=\frac{264\pm\sqrt{69696-47472}}{24}=\frac{264\pm\sqrt{22224}}{24}=\frac{264\pm149.08}{24}$. We get $x_1=\frac{264 + 149.08}{24}\approx17.96$ (rejected since $2x>23$) and $x_2=\frac{264-149.08}{24}\approx4.8$. Substitute $x = 4.8$ into $V(x)$: $V(4.8)=(43-2\times4.8)(23 - 2\times4.8)\times4.8=(43 - 9.6)(23 - 9.6)\times4.8=33.4\times13.4\times4.8\approx2028.07$.

Step4: Find volume formula for part b

The length and width of the box are $s - 2x$ and the height is $x$, so $V(x)=(s - 2x)^{2}x=x(s^{2}-4sx + 4x^{2})=4x^{3}-4sx^{2}+s^{2}x$.

Step5: Differentiate the volume function for part b

$V'(x)=12x^{2}-8sx + s^{2}$.

Step6: Set derivative equal to zero

Using the quadratic - formula with $a = 12$, $b=-8s$, $c = s^{2}$. $x=\frac{8s\pm\sqrt{64s^{2}-48s^{2}}}{24}=\frac{8s\pm\sqrt{16s^{2}}}{24}=\frac{8s\pm4s}{24}$. We get $x_1=\frac{s}{2}$ (rejected since $2x>s$) and $x_2=\frac{s}{6}$. Substitute $x=\frac{s}{6}$ into $V(x)$: $V(\frac{s}{6})=(s - 2\times\frac{s}{6})^{2}\times\frac{s}{6}=(\frac{2s}{3})^{2}\times\frac{s}{6}=\frac{4s^{2}}{9}\times\frac{s}{6}=\frac{2s^{3}}{27}$.