Answer :
If the dimensions of the cardboard sheet is 16 inches by 10 inches , then the maximum volume of the box is 144 in³ .
In the question ,
it is given that the length of the sheet is = 16 inch
and width of the sheet is = 10 inch
let width of box be ⇒ "w" ,
let length of box be ⇒ "l"
let the length of the corner cut out be = "x"
let volume of box be ⇒ "V" ;
we need to find the point of dV/dx that is maximum ,
given width = 16 ; and length = 10 ,
we have : width = x + w + x ⇒ w = 2(8 - x)
length = x + l + x = 10 ⇒ l = 2(5 - x)
the Volume of rectangular box is written as ; V = w*l*x ;
substituting the values of w , l and x ,
we get ;
V = 2(8 - x)2(5 - x)x
V = 4x³ - 52x² + 160x
differentiating with respect to "x" , and for critical point equating it to 0 ,
we have ;
12x² - 104x + 160 = 0
solving the above quadratic, we get ;
x = 2 & x = 20/3 .
On substituting the values , in d²V/dx² = 24x - 104 < 0 for x = 2 ,
that is maximum ;
Substituting x = 2 , in the V = 4x³ - 52x² + 160x ,
we get ;
Volume of the box as 144 in³ .
Therefore , the maximum volume of the box is = 144 in³ .
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