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A rectangular metal tank with an open top is to hold 13.5 cubic feet of liquid. What are the dimensions of the tank that require the least material to build?

Respuesta :

The volume is 13.5=xyz

and the surface area is:

s=xy+2xz+2yz

from the equation we get:

z=13.5/xy; k=13.5

z=k/xy

I will use k instead of a number just for the convenience we plug this into the second equation and we get:

s=xy+2k1/k+2k1/y

To find the minimum of this function we have to find the zeros of its first derivative.

sx will denote the first derivative with respect to x and sy will denote the first derivative with respect to sy.

[tex]sx = y - 2k \frac{1}{x {}^{2} } [/tex]

[tex]sy = x - 2k \frac{1}{y {}^{2} } [/tex]

Now let both derivatives go to zero and solve the system (this will give us the so called critical points)

[tex]0 = y - 2k \frac{1}{x {}^{2} } [/tex]

[tex]0 = x - 2k \frac{1}{y {}^{2} } [/tex]

[tex]y = 2k \frac{1}{x {}^{2} } [/tex]

[tex]x = 2k \frac{1}{y {}^{2} } [/tex]

Now, we plug the first equation into the other and we get:

[tex]x = \frac{ \frac{2k}{1} }{ \frac{4k {}^{2} }{x {}^{4} } } [/tex]

[tex]x {}^{3} = 2k [/tex]

[tex]x = (2.13.5) {}^{ \frac{1}{3} } [/tex]

x=3

Now we can calculate y:

[tex]y = 2k \frac{1}{x {}^{2} } [/tex]

[tex]y = 2.13.5 \frac{1}{3 {}^{3} } = 3 [/tex]

and finally we can calculate z:

[tex]z = \frac{13.5}{xy} [/tex]

[tex]z = \frac{13.5}{3.3} = 1.5[/tex]

and finally let's check our results

v=xyz=3×3×1.5=13.5

learn more about volume from here: https://brainly.com/question/28238693

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