Name: Second Derivative Test
Uploaded: 2019-07-19
Duration: 12 min 48 s
Description: The maxima and minima of a function are the largest and smallest value of the function. Watch this amazing video to find the maxima and minima of a function.

Question

The volume of the solid cuboid shown in the diagram is $576 c m^{3}$ and the surface area is $A c m^{2}$ . Find the minimum value of $A$ and state the dimensions of the cuboid for which this occurs.

Accepted Answer

Consider the following diagram,

The volume of this cuboid $= 576 c m^{3}$ , and the surface area of cuboid $= A c m^{2}$

We need to find the maximum value of $A$ , and the dimensions at which it occurs.

From the diagram, we can see that

The length $(l)$ of this cuboid $= 2 x c m$

Breadth $(b)$ of this cuboid $= x c m$

Height $(h)$ of this cuboid $= y c m$

We know that

Volume of cuboid $= length \times breadth \times height$

$\Rightarrow 576 = 2 x \times x \times y$

$\Rightarrow y = \frac{576}{2 x^{2}} = \frac{288}{x^{2}}$

We know that

Total Surface Area of cuboid $= 2 (l b + b h + h l)$

$\Rightarrow A = 2 (2 x \times x + x y + y \times 2 x)$

$= 2 (2 x^{2} + x \times \frac{288}{x^{2}} + \frac{288}{x^{2}} \times 2 x)$

$= 2 (2 x^{2} + \frac{288}{x} + \frac{576}{x})$

$= 2 (2 x^{2} + \frac{864}{x})$

$= 4 x^{2} + \frac{1728}{x}$

Now, $\frac{d A}{d x} = \frac{d}{d x} (4 x^{2} + \frac{1728}{x})$

$= \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (\frac{1728}{x})$

$= 4 \frac{d}{d x} (x^{2}) + 1728 \frac{d}{d x} (x^{- 1})$

$= 4 \times 2 x + 1728 \times (- 1) x^{- 2}$

$= 8 x - \frac{1728}{x^{2}}$

We know that the maximum or minimum value of the function occurs at stationary points. And, the point on the curve where the gradient is zero is called the stationary point.

$\frac{d A}{d x} = 8 x - \frac{1728}{x^{2}}$

And,

$\frac{d A}{d x} = 0$

$\Rightarrow 8 x^{3} - 1728 = 0$

$\Rightarrow x^{3} - 216 = 0$

$\Rightarrow (x - 6) (x^{2} + 6 x + 36) = 0$

$\Rightarrow x - 6 = 0$ or $x^{2} + 6 x + 36 = 0$

Since discriminant for $x^{2} + 6 x + 36 = 0$ is given by

$D = b^{2} - 4 a c = 36 - 4 \times 1 \times 36 < 0$

Thus, the equation $x^{2} + 6 x + 36 = 0$ has no roots.

Hence, stationary value of $x$ is $6$ .

Now, $\frac{d^{2} A}{d x^{2}} = \frac{d}{d x} (\frac{d A}{d x})$

$= \frac{d}{d x} (8 x - \frac{1728}{x^{2}})$

$= \frac{d}{d x} (8 x) - \frac{d}{d x} (1728 x^{- 2})$

$= 8 \frac{d}{d x} (x) - 1728 \frac{d}{d x} (x^{- 2})$

$= 8 - 1728 \times (- 2) x^{- 3}$

$= 8 + \frac{3456}{x^{3}}$

And, $\frac{d^{2} A}{d x^{2}}]_{x = 6} = 8 + \frac{3456}{{(6)}^{3}} = 24 > 0$

$\Rightarrow x = 6$ is a minimum point

Thus, we have

$A = 4 x^{2} + \frac{1728}{x}$

$= 4 {(6)}^{2} + \frac{1728}{6} = 144 + 288 = 432$

Hence, the minimum value of $A$ is $432 c m^{2}$ .

Since $x = 6 c m$ , which is the breadth of the cuboid

Length $= 2 x = 2 \times 6 = 12 c m$ . and height of the cuboid $= \frac{288}{6^{2}} = 8 c m$

Hence, the dimensions of the cuboid for which it has minimum area are $12 cm by 6 cm by 8 cm$ .

The volume of the solid cuboid shown in the diagram is $576 c m^{3}$ and the surface area is $A c m^{2}$ . Find the minimum value of $A$ and state the dimensions of the cuboid for which this occurs.

Important Questions on Further Differentiation

Learn and Prepare for any exam you want !

The volume of the solid cuboid shown in the diagram is 576cm3 and the surface area is Acm2. Find the minimum value of A and state the dimensions of the cuboid for which this occurs.

Important Questions on Further Differentiation

Learn and Prepare for any exam you want !

The volume of the solid cuboid shown in the diagram is $576 c m^{3}$ and the surface area is $A c m^{2}$ . Find the minimum value of $A$ and state the dimensions of the cuboid for which this occurs.