Question

(a) You have a 10 inch by 15 inch piece of tin which you plan to form into a box (without a top) by cutting a square from each corner and folding up the sides. How much should you cut from each corner so the resulting box has the greatest volume? (b) If the piece of tin is A inches by B inches, how much should you cut from each corner so the resulting box has the greatest volume?

Answer 1

Resulting box has the greatest volume for the values  (25 ± 5√7)/6 .

This is a problem that can be solved using derivatives , maxima & minima and common logic.

Hence , going by logic :

Creating a flap of 'a' inches in width, the base of the box will be

 (10 - 2a) by (15 - 2a)

and the depth of the box will be the width of the fold-up flap: a.

Then the volume of the box is

 v = $a(10 -2a)(15 -2a) = 150a -50a^2 +4a^3$

Using the derivative of the volume will be zero at the maximum volume.

 0 = $dv/da = 150 -100a +12a^2$

This has roots at

 a = (100 ±√(100² - 4(12)(150)))/(2·12)

 a = (100 ± √2800)/24 = (25 ± 5√7)/6

Only the smaller of these solutions gives a maximum volume.

You should cut (5/6)(5-√7) ≈ 1.962 inches to obtain the greatest volume.

Similarly , replacing the values of 10 by A and 15 by B , a generalized solution can be formed .

To know more about maxima and minima, go to brainly.com/question/29562544

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