Respuesta :
dimensions of the nickel plated square base: x*x = x^2
height: y
dimensions of the 4 silver plated sides: xy each
dimensions of the nickel plated top: x^2
Volume = 30cm^3 = yx^2 => y = 30 / x^2
Cost of the sides: 4 * xy * $3
Cost ot the top and the bottom: 2 * x^2 * $1
Total cost: 12xy + 2x^2
replace y by 30/x^2
=> cost = 12x * (30/x^2) + 2x^2 = 360 / x + 2x^2
Minimum cost => d [cost] / dx = 0 = - 360/x^2 + 4x =0
=> 90/x^2 - x =0
=> 90 - x^3 = 0
=> x^3 = 90
=> x = ∛(90) = 4.48
=> y = 30 / (4.48)^2 = 1.49
Answer: base: 4.48 cm* 4.48 cm; height: 1.49 cm
height: y
dimensions of the 4 silver plated sides: xy each
dimensions of the nickel plated top: x^2
Volume = 30cm^3 = yx^2 => y = 30 / x^2
Cost of the sides: 4 * xy * $3
Cost ot the top and the bottom: 2 * x^2 * $1
Total cost: 12xy + 2x^2
replace y by 30/x^2
=> cost = 12x * (30/x^2) + 2x^2 = 360 / x + 2x^2
Minimum cost => d [cost] / dx = 0 = - 360/x^2 + 4x =0
=> 90/x^2 - x =0
=> 90 - x^3 = 0
=> x^3 = 90
=> x = ∛(90) = 4.48
=> y = 30 / (4.48)^2 = 1.49
Answer: base: 4.48 cm* 4.48 cm; height: 1.49 cm
The dimension of the box to minimize the cost of the materials is [tex]4.48cm \times 1.49cm.[/tex]
According to the question, the volume of the jewelry box is [tex]30\;\rm cm^3[/tex].
Let the dimension of the base of the jewelry box be [tex]x\times x[/tex] and the height be [tex]y[/tex]. So,
[tex]Volume=x\times x\times y\\30 cm^3 =x\times x\times y[/tex]
[tex]y=\dfrac{30}{x^2}[/tex]------------eq [tex]1[/tex]
It has [tex]4[/tex] sides,
So, the Cost of the sides [tex]=4\times x\times y \times \$3[/tex]
As the box has square base so top and bottom so, the cost of nickel plating is [tex]=2\times x^2 \times \$1[/tex]
Now, [tex]Total\; cost=12xy+2x^2[/tex]
From eq [tex]1[/tex],
[tex]Total\; cost=12x\dfrac{30}{x^2}+2x^2\\Total\; cost = \dfrac{360}{x^2}+2x^2[/tex]
Minimum cost is evaluated as,
[tex]\frac{\partial(cost)}{\partial x}=0[/tex]
[tex]0=\frac{\partial(\dfrac{360}{x^2}+2x^2)}{\partial x}[/tex]
[tex]\dfrac{90}{x^2}-x=0[/tex]
[tex]90=x^3[/tex]
Simplify further-
[tex]x=\sqrt[3]{90}[/tex]
[tex]x=4.48\;cm[/tex]
[tex]y=\dfrac{30}{4.48^2}[/tex]
[tex]y=1.49[/tex]
Hence, the dimension of the box to minimize the cost of the materials is [tex]4.48cm \times 1.49cm.[/tex]
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