A line with a slope of -4 passes through the point

Mathematics

1 answer:

Viktor [21]1 year ago

4 0

Answer:

y = -4x - 22

Step-by-step explanation:

m = -4

using (-6,2)

Slope-intercept: y = mx + b

2 = -4(-6) + b

2 = 24 + b

b = -22

y = -4x - 22

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If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid

Artist 52 [7]

Check the picture below. Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm². Also, recall, the base is a square, thus, length = width = x.

$\bf \textit{volume of a rectangular prism}\\\\
V=lwh\quad 
\begin{cases}
l = length\\
w=width\\
h=height\\
-----\\
w=l=x
\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\
-------------------------------\\\\
\textit{surface area}\\\\
S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h
\\\\\\
\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\
-------------------------------\\\\
V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3$

so.. that'd be the V(x) for such box, now, where is the maximum point at?

$\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2
\\\\\\
\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100
\\\\\\
x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}$

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it. Check the second picture below.

so, the volume will then be at $\bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3$