Respuesta :
Answer:
x = 18/5
y = 18/5
z = 1.8
Step-by-step explanation:
Remember that the volume for the box is given by the formula
V(x,y,z) = xyz = 23.328
Since our cardboard has no lid then the formula for the surface area would be given by
A = xy + 2xz + 2yz.
The function that we want to optimize given the constrain of the Volume is A.
From the volume equation we find that
z = 23.328 / (xy)
we can plug that in A, and we get
[tex]A = xy + \frac{46.656x}{xy} + \frac{46.656y}{xy}\\[/tex]
If we equal each partial derivative to zero we get
[tex]y - \frac{46.656}{x^2} = 0 \\\\\x - \frac{46.656}{y^2} = 0[/tex]
From those equations we get
[tex]yx^2 = 46.656 \\xy^2 = 46.656[/tex]
If you solve that system of equations you get that
x = 18/5
y = 18/5
z = 1.8
Answer:
x,y,z=36cm, 36cm, 18cm
Step-by-step explanation:
Volume of the Cardboard =[tex]23,328 \:cm^3[/tex]
The dimensions of the box are x, y and z.
Volume of a Cuboid = xyz.
[tex]xyz=23328\\z=\dfrac{23328}{xy}[/tex]
Since the top is open, there is only one 'xy' side.
Surface Area, f(x,y,z)= xy+2yz+2xz
[tex]f(x,y)= xy+2y(\dfrac{23328}{xy})+2x(\dfrac{23328}{xy})\\f(x,y)= xy+2(\dfrac{23328}{x})+2(\dfrac{23328}{y})\\f(x,y)= xy+\dfrac{46656}{x}+\dfrac{46656}{y}[/tex]
We take the first derivative of f(x,y)
[tex]f_x(x,y)= y-\dfrac{46656}{x^2}\\f_y(x,y)= x-\dfrac{46656}{y^2}[/tex]
Equate the derivatives to zero and solve for x and y respectively.
[tex]y-\dfrac{46656}{x^2}=0\\y=\dfrac{46656}{x^2}\\y=46656x^{-2}\\Similarly\\x=\dfrac{46656}{y^2}\\x-\dfrac{46656}{y^2}\\x=46656y^{-2}[/tex]
Substituting x into y and vice versa
[tex]y=46656(46656y^{-2})^{-2}=46656^{-1}y^{4}\\y=46656^{-1}y^{4}\\y^3=46656\\y=36x=46656y^{-2}=46656(46656x^{-2})^{-2}=46656^{-1}x^{4}\\x=46656^{-1}x^{4}\\x^3=46656\\x=36[/tex]
Recall:
[tex]z=\dfrac{23328}{xy}\\z=\dfrac{23328}{36 X 36}=18\\[/tex]
x,y,z=36cm, 36cm, 18cm
