Answer:
The value of x would be ![\frac{5}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B3%7D)
Step-by-step explanation:
Given,
The dimension of the cardboard = 10 ft by 10 ft,
∵ After removing four equal squares of size x ( in ft ) from the corners,
The dimension of the resultant box would be,
Length = ( 10 - 2x ) ft,
Width = ( 10 - 2x ) ft,
Height = x ft,
The volume of box,
![V=(10-2x)\times (10 - 2x)\times x=x(10-2x)^2 = x(100 - 40x + 4x^2)=100x - 40x^2 + 4x^3](https://tex.z-dn.net/?f=V%3D%2810-2x%29%5Ctimes%20%2810%20-%202x%29%5Ctimes%20x%3Dx%2810-2x%29%5E2%20%3D%20x%28100%20-%2040x%20%2B%204x%5E2%29%3D100x%20-%2040x%5E2%20%2B%204x%5E3)
Differentiating with respect to x,
![V'=100 - 80x + 12x^2](https://tex.z-dn.net/?f=V%27%3D100%20-%2080x%20%2B%2012x%5E2)
Again differentiating with respect to x,
![V''=-80 + 24x](https://tex.z-dn.net/?f=V%27%27%3D-80%20%2B%2024x)
For maxima or minima,
![V'=0](https://tex.z-dn.net/?f=V%27%3D0)
![\implies 100 - 80x + 12x^2 = 0](https://tex.z-dn.net/?f=%5Cimplies%20100%20-%2080x%20%2B%2012x%5E2%20%3D%200)
![\implies 3x^2 - 20x + 25=0](https://tex.z-dn.net/?f=%5Cimplies%203x%5E2%20-%2020x%20%2B%2025%3D0)
By quadratic formula,
![x=\frac{20\pm \sqrt{20^2-4\times 3\times 25}}{6}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B20%5Cpm%20%5Csqrt%7B20%5E2-4%5Ctimes%203%5Ctimes%2025%7D%7D%7B6%7D)
![x=\frac{20\pm \sqrt{400 - 300}}{6}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B20%5Cpm%20%5Csqrt%7B400%20-%20300%7D%7D%7B6%7D)
![x=\frac{20\pm \sqrt{100}}{6}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B20%5Cpm%20%5Csqrt%7B100%7D%7D%7B6%7D)
![x=\frac{20\pm 10}{6}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B20%5Cpm%2010%7D%7B6%7D)
![\implies x = \frac{10}{6}=\frac{5}{3}\text{ or } x = 5](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%20%5Cfrac%7B10%7D%7B6%7D%3D%5Cfrac%7B5%7D%7B3%7D%5Ctext%7B%20or%20%7D%20x%20%3D%205)
For x = 5/3, V'' = negative,
While for x = 5, V'' = Positive,
Hence, the value of x would be 5/3 ft for maximising the volume.