Respuesta :
Answer:
The value of x would be [tex]\frac{5}{3}[/tex]
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,
[tex]V=(10-2x)\times (10 - 2x)\times x=x(10-2x)^2 = x(100 - 40x + 4x^2)=100x - 40x^2 + 4x^3[/tex]
Differentiating with respect to x,
[tex]V'=100 - 80x + 12x^2[/tex]
Again differentiating with respect to x,
[tex]V''=-80 + 24x[/tex]
For maxima or minima,
[tex]V'=0[/tex]
[tex]\implies 100 - 80x + 12x^2 = 0[/tex]
[tex]\implies 3x^2 - 20x + 25=0[/tex]
By quadratic formula,
[tex]x=\frac{20\pm \sqrt{20^2-4\times 3\times 25}}{6}[/tex]
[tex]x=\frac{20\pm \sqrt{400 - 300}}{6}[/tex]
[tex]x=\frac{20\pm \sqrt{100}}{6}[/tex]
[tex]x=\frac{20\pm 10}{6}[/tex]
[tex]\implies x = \frac{10}{6}=\frac{5}{3}\text{ or } x = 5[/tex]
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.
