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3 years ago

7

If 1900 square centimeters of material are available to make a box with a square base and an open top, find the largest possible

volume of the box.

1 answer:

Evgen [1.6K]3 years ago

3 0

Answer:

Volume = $7969$ cubic centimeter

Step-by-step explanation:

Let the length of each side of the base of the box be A and the height of the box be H.

Area of material required to make the box is equal to is $A^2 + 4*A*H.$

$A^2 + 4*A*H = 1900$

Rearranging the above equation, we get -

$`H = \frac{(1900 - A^2)}{(4*A)}$

Volume of box is equal to product of base area of box and the height of the box -

$V = A*A* H$

Substituting the given area we get -

$\frac{A^2*(1900 - A^2)}{4A} = \frac{(1900*A - A^3)}{4}$

For maximum volume

$\frac{dV}{dA} =0$

$\frac{ 1900}{4} - \frac{3*A^2}{4} = 0$

$A^2 = \frac{1900}{3}$

Volume of the box

= $\frac{\frac{1900}{3}*(1900 - \frac{1900}{3}) }{4 * \sqrt{\frac{1900}{3} } }$

= $7969$ cubic centimeter

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3 years ago

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

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The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

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Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

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$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

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3 years ago

Lauren is sewing a quilt using fabric squares.

KengaRu [80]

Answer:

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