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Oksana_A [137]
3 years ago
5

The volume of a cylinder is 4πx3 cubic units and its height is x units. Which expression represents the radius of the cylinder,

in units?
Mathematics
1 answer:
olga55 [171]3 years ago
4 0
\bf \textit{volume of a cylinder}\\\\
V=\pi r^2 h\qquad 
\begin{cases}
r=radius\\
h=height\\
-----\\
V=4\pi x^3\\
h=x
\end{cases}\implies 4\pi x^3=\pi \cdot r^2\cdot x
\\\\\\
\cfrac{4\pi x^3}{\pi x}=r^2\implies \implies 2^2x^2=r^2\implies \sqrt{2^2x^2}=r\implies 2x=r
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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi
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Answer:

  • Base Length of 84cm
  • Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, V=x^2h=296352

h=\dfrac{296352}{x^2}

Surface Area of the box = Base Area + Area of 4 sides

A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}

Step 2: Find the derivative of A(x)

If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}

Step 3: Set A'(x)=0 and solve for x

A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42

Therefore, the dimensions that minimizes the box surface area are:

  • Base Length of 84cm
  • Height of 42 cm.
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