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A closed box is constructed of 4200cm2 of cardboard. The box is a cuboid, with height hcm and square base of side xcm . What is the value of x which maximises the volume of the box? Give your answer in cm correct to 3 significant figures. [Answer format: 00.0 cm]

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thaovtp1407

Answer:

x is 10[tex]\sqrt{7}[/tex]

Step-by-step explanation:

  • Let x is the side of the base
  • Let h is the height

Given that:

  • The area A = 4200 cm2

<=> 2[tex]x^{2}[/tex] + 4xh = 4200 cm2

<=> 4xh = 4200 - 2[tex]x^{2}[/tex]  

<=> h = (4200 - 2[tex]x^{2}[/tex]  )/4x

  • The volume of the box:

V = [tex]x^{2}[/tex]  h

<=> V = [tex]x^{2}[/tex]  (4200 - 2[tex]x^{2}[/tex]  )/4x

<=> V = (4200x - 2[tex]x^{3}[/tex] )/4

  • To find the maximum volume, we differentiate volume with respect to x:

dV/dx = (4200 - 6[tex]x^{2}[/tex] )/4

Set dV/dx = 0, we have:

4200 - 6[tex]x^{2}[/tex] =0

<=> [tex]x^{2}[/tex]  = 700

<=> x = 10[tex]\sqrt{7}[/tex]

  • We differentiate again

d²V/dx² =  -12x/4

It is negative so the volume is maximum.

So x is 10[tex]\sqrt{7}[/tex]

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