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  • A metal box with a square base and vertical sides is to contain 1024 cm3. The material for the top and bottom costs Rs 5/cm2 and the material for the sides costs Rs 2.50/cm2. Find the least cost of the box.

A metal box with a square base and vertical sides is to contain 1024 cm^3. The material for the top and bottom costs Rs 5/cm^2 and the material for the sides costs Rs 2.50/cm^2. Find the least cost of the box.

Answers (1)

Given: A metal box with a square base and vertical sides is to contain 1024 cm^3. The material for the top and bottom costs Rs 5/cm^2 and the material for the sides costs Rs 2.50/cm^2

To find: the minimum cost of the box

.

Take x cm as the side of the square

Take y cm as the vertical side of the metal box

According to the given information in the question, the formula used volume for square base is

V=base × height

Due to its square base, the formula of the volume is

V=x^2y
This is equal to $1024 \mathrm{~cm}^{3} . So, volume becomes
\\1024 \mathrm{~cm}^{3}=\mathrm{x}^{2} \mathrm{y}$\\ $\mathrm{y}=\frac{1024}{\mathrm{x}^{2}} \ldots \ldots$ (i)

Then we need to calculate the total area of the metal box.

Area of top and bottom = 2x\textsuperscript{2}cm\textsuperscript{2}

The mentioned material for the top and bottom costs Rs 5/cm\textsuperscript{2}, thus, the material cost for top and bottom becomes

Cost of top and bottom =Rs. 5(2x^2)

Area of one side of the metal box = xy cm

The total sides present in the metal box are 4, so

Thus, the total area of all the sides of the metal box = 4xy

The cost of the material for sides is Rs 2.50/cm^2

∴ Cost of all the sides of the metal box =Rs. 2.50(4xy)

The overall area of the metal box will be

A=2x\textsuperscript{2}+4xy

This will make the cost of the box to be

C=5(2x\textsuperscript{2}) + 2.50(4xy)

Putting the value of y from equation (i) in the above equation,

\\C=5\left(2 x^{2}\right)+2.50\left(4 x\left(\frac{1024}{x^{2}}\right)\right)$\\ $\Rightarrow C=10 x^{2}+\frac{10240}{x}$
Both the sides are differentiated with respect to x
\Rightarrow C^{\prime}=\frac{d\left(10 x^{2}+\frac{10240}{x}\right)}{d x}$
Using differentiation rule of sum, we get
\Rightarrow C^{\prime}=\frac{d\left(10 x^{2}\right)}{d x}+\frac{d\left(\frac{10240}{x}\right)}{d x}$

\Rightarrow C^{\prime}=10 \frac{d\left(x^{2}\right)}{d x}+10240 \frac{d\left(x^{-1}\right)}{d x}$
Then using the derivative, we get
\\\Rightarrow C^{\prime}=10 \times 2 x+10240 \times\left(-x^{-2}\right)$ \\$\Rightarrow C^{\prime}=20 x-\frac{10240}{x^{2}} \ldots$ (ii)
Take c=0 to find the minimum value of x by apply second derivative test, so the above equation is equated with 0
\\20 x-\frac{10240}{x^{2}}=0$ \\\\$\Rightarrow 20 \mathrm{x}=\frac{10240}{\mathrm{x}^{2}}$ \\\\$\Rightarrow \mathrm{x}^{3}=\frac{10240}{20}$
$\Rightarrow x^{3}=512$
Solving this we get
$\Rightarrow x=8$
Again, differentiating equation (ii) with respect to x,
\Rightarrow C^{\prime \prime}=\frac{d\left(20 x-\frac{10240}{x^{2}}\right)}{d x}$
Using the differentiation rule of sum,
\\\Rightarrow C^{\prime \prime}=\frac{d(20 x)}{d x}-\frac{d\left(\frac{10240}{x^{2}}\right)}{d x}$ \\$\Rightarrow \mathrm{C}^{\prime \prime}=20 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}-10240 \frac{\mathrm{d}\left(\mathrm{x}^{-2}\right)}{\mathrm{dx}}$ \\$\Rightarrow \mathrm{C}^{\prime \prime}=20+20480 \times\left(-\mathrm{x}^{-3}\right)$ \\$\Rightarrow C^{\prime \prime}=20+\frac{20480}{x^{3}}$

At x=8, the above equation becomes,

\\\Rightarrow\left(C^{\prime \prime}\right)_{x=8}=20+\frac{20480}{(8)^{3}}$ \\$\Rightarrow\left(C^{\prime \prime}\right)_{x=8}=20+\frac{20480}{512}>0$
Now at x=8, C^{\prime}(8)=0$ and $C^{\prime \prime}(8)>0, so as per the second derivative test, x is a point of local minima and $\mathrm{C}(8)$ will be minimum value of C.
Hence least cost becomes
\\C_{x=8}=10(8)^{2}+\frac{10240}{8}$ \\$\Rightarrow C_{y=8}=640+1280=RS.1920
Hence the least cost of the metal box is Rs. 1920

Posted by

infoexpert22

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