a. The shape of the box is a cuboid.
Mathematicaly, the volume of a cuboid is L * B * H
With respect to the given dimensions in the question, the volume of the box would be;
w * ( w + 12) * (27-w)
Since we have the value of the volume already, we can simply expand the brackets and equate to the given volume.
Hence;
w(w + 12)(27-w) = 4860
w( w(27-w) +12(27-w)) = 4860
w(27w - w^2 + 324 -12w) = 4860
w(27w -12w + 324 - w^2) = 4860
w(15w + 324 - w^2) = 4860
Thus, we have ;
15w^2 + 324w - w^3 = 4860
So the polynomial equation that can be used to get the dimension of the box is ;
b. We want to solve the equation above
15w^2 + 324w - w^3 = 4860
We solve the equation so as to get the values of w
Kindly note that because the polynomial equation is cubic, we shall be expecting three answers;
Thus, we proceed as follows;
Let us express the polynomial in the normal form as follows
-w^3 + 15w^2 + 324w - 4860 = 0
Let us now factorize;
-w^2( w -15) + 324( w - 15) = 0
(w - 15)( -w^2 + 324) = 0
Thus;
w-15 = 0 or 324 - w^2 = 0
Hence, w = 15 or w^2 = 324
So w = 15 or w = √324
w = 15 or ± 18
So the values of w are 15 , 18 or -18
c. The possible dimensions
We simply substite the value of w at each of the sides
The sides are ; w, 27-w and w + 12
at w = 15, we have
15 , 27-15 and 15 + 12 = 15, 12 and 27
At w = 18, we have
18, 27-18 and 18 + 12
18, 9 and 30
Lastly, we have w = -18
so we have ;
-18 , 27-(-18) and -18 + 12
-18, 45 and -6
d. Which set of dimensions is preferred
The dimension set 1 and 2 are preferred
i.e 18, 9 and 30 or 15,12 and 27
This is becasuse the last set of dimensions is not possible because the dimensions of a cuboid cannot be negative