Width=w
length=(40-2w)/2=20-w
A=w(40-w)
A=40w-w²
A'=40-2w=0
2w=40
w=20

20x20 is the size of the pen having the greatest area. this tells us the largest rectangle for a given perimeter is a square

8 0

3 years ago

How do you find the length of the curve x=3t−t3x=3t-t^3, y=3t2y=3t^2, where 0≤t≤3√0<=t<=sqrt(3) ?

taurus [48]

It's simple, you just have to do this:

$L=\int\limits_{a}^{b}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt$

$x=3t-t^3$

$\frac{dx}{dt}=3-3t^2$

$y=3t^2$

$\frac{dy}{dt}=6t$

replacing

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{\left(3-3t^2\right)^2+\left(6t\right)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9-18t^2+9t^4+36t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{9+9t^4+18t^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}\sqrt{(3t^2+3)^2}~dt$

$L=\int\limits_{0}^{\sqrt{3}}(3t^2+3)~dt$

$L=t^3+3t|_0^{\sqrt{3}}$

$\boxed{\boxed{L=3\sqrt{3}+3\sqrt{3}=6\sqrt{3}}}$

8 0

3 years ago

Solve negative one fifth times the quantity 4 minus 14 times the quantity one fourth squared end quantity end quantity.

asambeis [7]

The value of the expression is one eight

<h3>How to solve the expression?</h3>

The expression is given as:

negative one fifth times the quantity 4 minus 14 times the quantity one fourth squared end quantity

Rewrite properly as:

-1/5 * (4 - 14) * (1/4)^2

Evaluate the difference

-1/5 * (-10) * (1/4)^2

Evaluate the exponent

-1/5 * (-10) * 1/16

This gives

2 * 1/16

Further, evaluate

1/8

Hence, the value of the expression is one eight