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Math Help - Applications of differentiation : Max and Minimization

  1. #1
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    Applications of differentiation : Max and Minimization

    (1)A rectangular tank with no top and a volume of 8m3 is to be made from a sheet of metal. The width of the base is 4m. Show that the surface area is given by
    S = 4x + 4 + 16/x , where x is the length of the base and find the dimensions of the tank so that minimum material is used.

    (2) The breaking weight of a cantilever beam is directly proportional to b*d^2, where b is the breadth of the beam and d is its depth. Find the cross-section dimensions of a beam with a cross-section perimeter of 60cm that has maximum breaking weight.

    (3)An oil tank, designed to hold 10 m3, is in the form of a cylinder with hemispherical top.
    (i) Show that the surface area (A) of the tank is given by A = 20/r + 5/3 * πr^2 , where r is the radius and h is height of the cylinder.
    (ii) Find the dimensions (r and h) of the tank that has least surface area and calculate the least surface area.


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  2. #2
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    Maxima and minima of differentiable functions are always attained at either boundary points (the largest or smallest value of x possible, for instance) or points at which the tangent line to the curve is horizontal, i.e., f'(x)\,=\,0.

    (1) To find the surface area S of a tank in terms of its length x, we begin by writing down what we know:

    \begin{aligned}<br />
wxh&=8\,\mbox{m}^3\\<br />
wx+2wh+2xh&=S.<br />
\end{aligned}

    Since w=4, we have

    \begin{aligned}<br />
4xh&=8\,\mbox{m}^3\\<br />
h&=\frac{2}{x}\,\mbox{m}^3.<br />
\end{aligned}

    All that remains is to substitute these expressions in the first equation. To find the value of x that allows us minimum surface area, we look for points at which S'(x)=0:

    \begin{aligned}<br />
S'(x)&=\frac{d}{dx}\left(4x+\frac{16}{x}+4\right)\  \<br />
&=\frac{d}{dx}(4x)+\frac{d}{dx}\left(\frac{16}{x}\  right)+\frac{d}{dx}(4)\\<br />
&=4-\frac{16}{x^2}+0\\<br />
&=4-\frac{16}{x^2}=0.<br />
\end{aligned}<br />

    As we know that S is always positive and that the surface area would tend to \infty as x approached 0 or \infty, we can be sure that the mininum is not attained at a boundary point. All we must do, then, is find the solution of S'(x)=0.


    (3) Here's a hint:

    \begin{aligned}<br />
S&=2\pi rh + \frac{1}{2}\cdot4\pi r^2 + \pi r^2\;\;\;\;\;\;\;\;\;\;\mbox{(cylinder + half sphere + base)}\\<br />
A&=\pi r^2h + \frac{1}{2}\cdot\frac{4}{3}\pi r^3 = 10\,\mbox{m}^3.<br />
\end{aligned}

    To find S in terms of r, we may begin by finding h in terms of r.

    Hope this helps!
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