When the surrounding flaps are folded up, the base of the box will have dimensions $11-2x$ by $16-2x$ , and the box will have a height of $x$ . So the box has volume, as a function of $x$ ,

$V(x)=(11-2x)(16-2x)x=176x-54x^2+4x^3$

I don't know what technology is available to you, but we can determine an exact value for $x$ that maximizes the volume by using calculus.

Differentiating $V$ with respect to $x$ gives

$\dfrac{\mathrm dV}{\mathrm dx}=176-108x+12x^2$

and setting this equal to 0 gives two critical points at

$x=\dfrac{27\pm\sqrt{201}}6\implies x\approx2.1\text{ or }x\approx6.9$

For the larger critical point we would get a negative volume, so we ignore that one. Then the largest volume would be about 168.5 cubic in.

7 0

3 years ago

Solve the open sentence. –2 ≤ n + 4 ≤ 7

bekas [8.4K]

The answer is n ≥ –6 and n ≤ 3 .hope this helps

6 0

3 years ago

Read 2 more answers

Use the area model to write a fraction equivalent to <br> 1/5

juin [17]

A fraction equivalent to 1/5 could be 2/10

7 0

3 years ago