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2 years ago

Solve using quadratic formula. 7x^2-12x+3=0

1 answer:

4 0

X = (12 +- sqrt (144 -4(7)(3))/14
x = (12 + - sqrt (144 - 84))/14
x = (12 + - sqrt(60))/14
x = (12 + - 2sqrt 15))/14
x = (6 + - sqrt 15) / 7

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The dimensions that minimize the perimeter are $\mathbf{10\sqrt{10 },10\sqrt{10 }}$

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$\mathbf{A = xy = 1000}$

Make x the subject

$\mathbf{x = \frac{1000}{y}}$

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$\mathbf{P = 2(x + y)}$

Substitute $\mathbf{x = \frac{1000}{y}}$

$\mathbf{P = 2(\frac{1000}{y} + y)}$

Expand

$\mathbf{P = \frac{2000}{y} + 2y}$

Differentiate

$\mathbf{P' = -\frac{2000}{y^2} + 2}$

Set to 0

$\mathbf{ -\frac{2000}{y^2} + 2 = 0}$

Rewrite as:

$\mathbf{ -\frac{2000}{y^2} = -2}$

Divide both sides by -1

$\mathbf{\frac{2000}{y^2} = 2}$

Multiply y^2

$\mathbf{2000 = 2y^2}$

Divide by 2

$\mathbf{1000 = y^2}$

Take square roots of both sides

$\mathbf{y = \sqrt{1000 }}$

$\mathbf{y = 10\sqrt{10 }}$

Substitute $\mathbf{y = \sqrt{1000 }}$ in $\mathbf{x = \frac{1000}{y}}$

$\mathbf{x = \frac{1000}{\sqrt{1000}}}$

$\mathbf{x = \sqrt{1000}}$

$\mathbf{x = 10\sqrt{10 }}$

Hence, the dimensions that minimize the perimeter are $\mathbf{10\sqrt{10 },10\sqrt{10 }}$