Variable y varies directly with x2, and y = 96 when x = 4.

1 answer:

4 0

Direct variation is y = kx, where k is the constant of variation.
But now it says y varies directly with x2 (or 2x), so now the x in the equation is 2x.

The equation is y = k(2x)
Now you find k.
y = 96 when x = 4.
(96) = k(2*4)
96 = k(8)
k = 12

The equation is now y = 12(2x)
To find the value of y when x=2, plug 2 into the equation you made. y = 12(2*2) y = 48
_________________
Now it's with a "quadratic variation," which is the same thing except x is squared.
The equation is y = kx^2

But y varies directly with x2 (same thing as 2x), so now it's y = k(2x)^2.

Now you find k by substituting y and x values that were given.
y = 180 when x = 6
(180) = k(2*6)^2
180 = k(12)^2
180 = k(144)
k = 1.25
k, 1.25, is the constant of variation.

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now $\bf \begin{cases}
A=l\cdot w\\\\\
A=3000000\\
----------\\
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\frac{3000000}{w}=l
\end{cases}\qquad thus
\\\\\\
P=3w+2l\implies P=3w+2\left( \cfrac{3000000}{w} \right)\implies P=3w+\cfrac{6000000}{w}
\\\\\\
P=3w+6000000w^{-1}\\\\
-----------------------------\\\\
now\qquad \cfrac{dP}{dw}=3-\cfrac{6000000}{w^2}\implies \cfrac{dP}{dw}=\cfrac{3w^2-6000000}{w^2}
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\textit{so the critical points are at }
\begin{cases}
0=w^2\\\\
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\end{cases}$

solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum

notice the $0=w^2\implies 0=w$ so. you can pretty much skip that one, though is a valid critical point, the width can't clearly be 0

so.. check the critical points on the other