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an open-top box is to be made by cutting small congruent squares from the corners of a 12- in.-by-12-in. sheet of tin and bending up the sides. how large should the squares cut from the corners be to make the box hold as much as possible? what are the dimensions of the box with the largest volume?

Respuesta :

The squares cut from the corners should be large in width and length 6 to hold the box as much as possible. The dimensions of the box with the largest volume are 6 x 6.

If the squares cut from the corners are h × h inches

The open-top box will have

A height of h

A width of 12 - 2h

and a length of 12 - 2h

So its volume will be

V(h) = h × (12 - 2h) × (12 - 2h)

= 4h³ - 48h² + 144h    (square inches)

dV/dh= 12h² - 96h + 144

Critical points occur when the derivative (dV/dh) is zero.

12h² - 96h + 144 = 0

h² - 8h + 12 = 0

(h − 2)(h − 6) = 0

It is clear that h = 6 is not the crucial point for the maximum as it would produce widths and lengths of zero (and so a volume of zero).

In order to cut out squares measuring 6 × 6 inches from the corners of the bigger sheet, the most volume may be obtained.

To learn more about volume

https://brainly.com/question/1578538

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