Question

Amongst all open ( from the top) right circular cylindrical boxes of volume 125π cm³. find the dimensions of the box which has the least surface area?

Answer 1

The dimensions of the box which has the least surface are radius = 5 cm and height = 5 cm .

Given that the volume of the right circular cylindrical box = 125π cm³.

Let the radius of the cylinder be r and the height be equal to h.

=> Volume = πr²h

=> πr²h = 125π

=> r²h = 125

=> h = 125/r²

Surface area of the box = 2πrh + πr²

= 2π r × (125/r²) + πr²

= (250π/r) + πr²

Let f(r) = (250π/r) + πr² be a function w.r.t r.

Differentiating f(r) w.r.t r to find the point of minima .

f'(r) = (-250π/r²) + 2πr = 0

=> 2πr = 250π/r²

=> r³ = 125

=> r = 5 cm

h = 125/25 = 5 cm

The dimension of the cylindrical box is radius = 5 cm and height = 5 cm .

Answer 2

r=  3.97cm and h=  7.93 cm

Step-by-step explanation:

Given V= 125π cm³

Volume of right circular cylinder = πr$r^{2}$h

Surface area of can = 2 πrh + 2π$r^{2}$ …….. (i)

Put the value of h in (i), we get

$\begin{aligned}&=2 \pi r \frac{125}{r^{2}}+2 \pi r^{2}\\&s(r)=\frac{250 \pi}{r}+2 \pi \mathbf{r}^{2}\\&\Rightarrow s^{\prime}(r)=-\frac{250 \pi}{r^{2}}+4 \pi \mathbf{r}\\&\text { and } s^{\prime \prime}(r)=-\frac{2 \times 250 \pi}{r^{3}}+4 \pi\end{aligned}$

$\begin{aligned}&\text { If } s^{\prime}(r)=0 \text { then }-\frac{250 \pi}{r^{2}}+4 \pi r=0\\&4 \pi r=\frac{250 \pi}{r^{2}}\\&r=3.9685 \approx 3.97\end{aligned}$

$\begin{aligned}&\text { Now } h=\frac{125}{(3.97)^{2}}\\&h=7.93 \mathrm{cm}\end{aligned}$

Therefore minimum surface area can have radius 3.97 and height 7.93 cm

To learn more

(i)A solid consisting of a right circular cone standing on a hemisphere, is placed upright in a right circular cylinder full of water and touches the bottom

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