Respuesta :
It is given in the question that
A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by
[tex]V = x^2h cm^3[/tex]
And the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min.
Here we differentiate V with respect to t, and we use product rule, that is
[tex]V = 2xh(dx/dt)+ x^2(dh/dt)[/tex]
Substituting the given values , we will get
[tex][tex]dV/dt = 2(12)(6)(-2)+ 12^2(1)[/tex][/tex]
[tex]dV/dt = -288+144 = -144cm^3/min[/tex]
So at that moment, the volume is decreasing at the rate of
[tex]144 cm^3/min[/tex]
The volume is decreasing at the rate of [tex]\boxed{{\mathbf{144c}}{{\mathbf{m}}^{\mathbf{3}}}{\mathbf{/min}}}[/tex] .
Further explanation:
Given:
It is given that a rectangular box a square base with a length of the side is [tex]x{\text{ cm}}[/tex] and has a height of [tex]h{\text{ cm}}[/tex] .
Step by step explanation:
Step 1:
The volume of the rectangular box can be calculated as,
[tex]V=lbh[/tex]
Here, [tex]l[/tex] is the length, [tex]b[/tex] is breadth, [tex]h[/tex] is the height.
Therefore, volume of the box can be calculated as,
[tex]V={x^2}h{\text{ c}}{{\text{m}}^{\text{3}}}[/tex]
Step 2:
Now differentiate volume with respect to [tex]l[/tex] both sides by the use of product rule.
[tex]\frac{{dV}}{{dt}}=2xh\left({\frac{{dx}}{{dt}}}\right)+{x^2}\left({\frac{{dh}}{{dt}}}\right)[/tex]
It is given that the length of the base is [tex]12{\text{ cm}}[/tex] and it decreasing at the rate of .
The height of the box is [tex]6{\text{ cm}}[/tex] and the height is increasing at the rate of [tex]1{\text{ cm/min}}[/tex] .
Therefore, the values are [tex]x=12,h=6,\frac{{dx}}{{dt}}=-2{\text{ and}}\frac{{dh}}{{dt}}=1[/tex]
Now substitute the values [tex]x=12,h=6,\frac{{dx}}{{dt}}=-2{\text{ and}}\frac{{dh}}{{dt}}=1[/tex] in the differential equation.
[tex]\begin{gathered}\frac{{dV}}{{dt}}=2xh\left({\frac{{dx}}{{dt}}}\right)+{x^2}\left({\frac{{dh}}{{dt}}}\right)\hfill\\\frac{{dV}}{{dt}}=2\left({12}\right)\left(6\right)\left({-2}\right)+{\left({12}\right)^2}\left(1\right)\hfill\\\frac{{dV}}{{dt}}=-144{\text{c}}{{\text{m}}^{\text{3}}}/\min\hfill\\\end{gathered}[/tex]
Therefore, it can be seen that the volume is decreasing as the value of [tex]\frac{{dV}}{{dt}}[/tex] involves negative sign.
Thus, the volume is decreasing at the rate of [tex]144c{m^3}/\min[/tex] .
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Answer details:
Grade: Middle school
Subject: Mathematics
Chapter: Application of derivative
Keywords: side, length, square, base, differentiate, decreasing, increasing, rectangular box, respect, product rule, height, rate, volume of the box
