Answer:
Dimension of the box is ![16.1\times 7.1\times 2.45](https://tex.z-dn.net/?f=16.1%5Ctimes%207.1%5Ctimes%202.45)
The volume of the box is 280.05 in³.
Step-by-step explanation:
Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.
To find : The dimensions and the volume of the box?
Solution :
Let h be the height of the box which is the side length of a corner square.
According to question,
A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.
The length of the box is ![L=21-2h](https://tex.z-dn.net/?f=L%3D21-2h)
The width of the box is ![W=12-2h](https://tex.z-dn.net/?f=W%3D12-2h)
The volume of the box is ![V=L\times W\times H](https://tex.z-dn.net/?f=V%3DL%5Ctimes%20W%5Ctimes%20H)
![V=(21-2h)\times (12-2h)\times h](https://tex.z-dn.net/?f=V%3D%2821-2h%29%5Ctimes%20%2812-2h%29%5Ctimes%20h)
![V=(21-2h)\times (12h-2h^2)](https://tex.z-dn.net/?f=V%3D%2821-2h%29%5Ctimes%20%2812h-2h%5E2%29)
![V=252h-42h^2-24h^2+4h^3](https://tex.z-dn.net/?f=V%3D252h-42h%5E2-24h%5E2%2B4h%5E3)
![V=4h^3-66h^2+252h](https://tex.z-dn.net/?f=V%3D4h%5E3-66h%5E2%2B252h)
To maximize the volume we find derivative of volume and put it to zero.
![V'=12h^2-132h+252](https://tex.z-dn.net/?f=V%27%3D12h%5E2-132h%2B252)
![0=12h^2-132h+252](https://tex.z-dn.net/?f=0%3D12h%5E2-132h%2B252)
Solving by quadratic formula,
![h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B-%28-132%29%5Cpm%5Csqrt%7B132%5E2-4%2812%29%28252%29%7D%7D%7B2%2812%29%7D)
![h=\frac{132\pm72.99}{24}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B132%5Cpm72.99%7D%7B24%7D)
![h=2.45,8.54](https://tex.z-dn.net/?f=h%3D2.45%2C8.54)
Now, substitute the value of h in the volume,
![V=4h^3-66h^2+252h](https://tex.z-dn.net/?f=V%3D4h%5E3-66h%5E2%2B252h)
When, h=2.45
![V=4(2.45)^3-66(2.45)^2+252(2.45)](https://tex.z-dn.net/?f=V%3D4%282.45%29%5E3-66%282.45%29%5E2%2B252%282.45%29)
![V\approx 280.05](https://tex.z-dn.net/?f=V%5Capprox%20280.05)
When, h=8.54
![V=4(8.54)^3-66(8.54)^2+252(8.54)](https://tex.z-dn.net/?f=V%3D4%288.54%29%5E3-66%288.54%29%5E2%2B252%288.54%29)
![V\approx -170.06](https://tex.z-dn.net/?f=V%5Capprox%20-170.06)
Rejecting the negative volume as it is not possible.
Therefore, The volume of the box is 280.05 in³.
The dimension of the box is
The height of the box is h=2.45
The length of the box is ![L=21-2(2.45)=16.1](https://tex.z-dn.net/?f=L%3D21-2%282.45%29%3D16.1)
The width of the box is ![W=12-2(2.45)=7.1](https://tex.z-dn.net/?f=W%3D12-2%282.45%29%3D7.1)
So, Dimension of the box is ![16.1\times 7.1\times 2.45](https://tex.z-dn.net/?f=16.1%5Ctimes%207.1%5Ctimes%202.45)