A rectangular visiting card is to contain 24 sq.cm. of printed matter

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Q) A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be 1½ cm as shown below :

On the basis of the above information, answer the following questions :
(i) Write the expression for the area of the visiting card in terms of x.
(ii) Obtain the dimensions of the card of minimum area.

Ans:

(i) Expression of Visiting card Area:

Step 1: Since the length of printed matter is X and breadth is Y

Therefore, area of printed matter is XY

since it is given that the printed matter area is 24 sq cm

X Y = 24

∴ Y = $\frac{24}{\times}$ ………….(i)

Step 2: Length of Visiting card = (X + 3) cm

and breadth of visiting card = (Y + 2) cm

Hence, Area of visiting card, A = length x Breadth

∴ A = (X + 3) x (Y +2)

∴ A = X Y + 3 Y + 2 X + 6

Step 3: By substituting value of Y from equation (i), we get:

A = X ( $\frac{24}{\times}$ ) + 3( $\frac{24}{\times}$ ) + 2 X + 6

∴ A = 24 + $\frac{72}{\times}$ + 2 X + 6

∴ A = 2 X + $\frac{72}{\times}$ + 30

(ii) dimensions of the card of minimum area:

Step 1: To get minimum Area, we need to take derivative of A w.r.t. X.

$\frac{d}{dx} A= \frac{d}{dx}(2x) + \frac{d}{dx} (\frac{72}{\times}) + \frac{d}{dx} (30)$

= $2 - (\frac{72}{\times^2})$

Step 2: We know that the minimum is obtained when the derivative is zero

∴ $\frac{d}{dx} A$ = 0

∴ 2 – $\frac{72}{\times^2}$ = 0

∴ $\frac{72}{\times^2}$ = 2

∴ 72 = 2 X²

∴ X² = 36

∴ X = 6 and X = -6

Here we reject X = – 6, because dimension can not be a negative number.

Hence X = 6 cm

and from equation (i), Y = $\frac{24}{6}$ = 4 cm

Dimensions of visiting card are:

Length = X + 3 = 6 + 3 = 9 cm

and breadth = Y + 2 = 4 + 2 = 6 cm

Therefore, the dimensions of the visiting card are: Length 9 cm and breadth 6 cm.

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