Answer:
[tex](a)[/tex] [tex]V = -H^3+ 5H^2 + 36H[/tex]
[tex](b)[/tex] [tex]Length = 5in; Width = 9in; Height = 5in[/tex] or [tex]Length = 3in; Width = 10in; Height = 6in[/tex]
Step-by-step explanation:
Let:
[tex]W=Width; L = Length; H = Height[/tex]
Given
[tex]W = 4 + H[/tex]
[tex]L = 9 - H[/tex]
Solving (a): The polynomial for the volume in terms of height
Volume (V) is calculated using:
[tex]V = L * W * H[/tex]
[tex]V = (9 - H) * (4 + H) * H[/tex]
Expand
[tex]V = (9 - H) * (4H + H^2)[/tex]
Open brackets
[tex]V = 36H + 9H^2 - 4H^2 - H^3[/tex]
[tex]V = 36H + 5H^2 - H^3[/tex]
Rewrite as:
[tex]V = -H^3+ 5H^2 + 36H[/tex]
Solving (b): The dimensions when Volume = 180
Substitute 180 for V in: [tex]V = -H^3+ 5H^2 + 36H[/tex]
[tex]180 = -H^3+ 5H^2 + 36H[/tex]
Rewrite as:
[tex]H^3 - 5H^2 - 36H + 180 = 0[/tex]
Factorize:
[tex]H^2(H - 5) - 36(H - 5) = 0[/tex]
[tex](H^2 - 36)(H - 5) = 0[/tex]
Express [tex]H^2 - 36[/tex] as difference of two squares
[tex](H - 6)(H + 6)(H - 5) =0[/tex]
This gives:
[tex]H - 6 =0==> H = 6[/tex]
[tex]H + 6 =0==> H = -6[/tex]
[tex]H - 5 =0==> H = 5[/tex]
Since the height can not be negative, then H = 5 or H = 6
Recall that:
[tex]W = 4 + H[/tex]
[tex]L = 9 - H[/tex]
When H = 5
[tex]W = 4 + 5 = 9[/tex]
[tex]L = 9 - 5 = 4[/tex]
When H = 6
[tex]W = 4 + 6 = 10[/tex]
[tex]L = 9 - 6 = 3[/tex]
Hence, the dimensions of the box are:
[tex]Length = 5in; Width = 9in; Height = 5in[/tex]
or
[tex]Length = 3in; Width = 10in; Height = 6in[/tex]
The volume of the box is given by -h³ + 5h² + 36h
Volume is the amount of space occupied by a three dimensional object. The volume (V) of a box is given by:
V = length * width * height
Let h represent the the height, hence width = h + 4, length = (9 - h)
V = h(h + 4)(9 - h) = h(5h -h² + 36) = -h³ + 5h² + 36h
For a volume of 180 in³:
180 = -h³ + 5h² + 36h
h = 5 or 6.
The volume of the box is given by -h³ + 5h² + 36h
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