Answer:
The height that make the surface area as small as possible is 10 cm
Step-by-step explanation:
Let
L -----> the length of the base of the box
W -----> the width of the base of the box
H ----> the height of the box
we know that
The volume of the box is equal to
[tex]V=LWH[/tex]
we have
[tex]V=4,500\ cm^3[/tex]
so
[tex]4,500=LWH[/tex] ----> equation A
[tex]L=2W[/tex] ------> equation B
substitute equation B in equation A
[tex]4,500=(2W)WH[/tex]
[tex]4,500=2W^2H[/tex]
Solve for H
[tex]H=\frac{2,250}{W^{2}}[/tex] -----> equation C
The surface area of a open box is equal to
[tex]SA=2(L+W)H+LW[/tex] ----> equation D
Substitute equation B and equation C in equation D
[tex]SA=2(2W+W)(\frac{2,250}{W^{2}})+(2W)W[/tex]
[tex]SA=6W(\frac{2,250}{W^{2}})+2W^2[/tex]
[tex]SA=6(\frac{2,250}{W})+2W^2[/tex]
[tex]SA=\frac{13,500}{W}+2W^2[/tex]
Convert to function notation
[tex]f(W)=\frac{13,500}{W}+2W^2[/tex]
Find the derivative
[tex]f'(W)=-\frac{13,500}{W^2}+4W[/tex]
set equal to zero
[tex]f'(W)=0[/tex]
[tex]0=-\frac{13,500}{W^2}+4W[/tex]
Solve for W
[tex]\frac{13,500}{W^2}=4W[/tex]
[tex]W^3=\frac{13,500}{4}[/tex]
[tex]W=15\ cm[/tex]
Find the value of L (equation B)
[tex]L=2(15)=30\ cm[/tex]
Find the value of H (equation C)
[tex]H=\frac{2,250}{15^{2}}=10\ cm[/tex]
therefore
The height that make the surface area as small as possible is 10 cm
