$y_1=ax+b$ is tangent to the curve $y_2=12-x^2,$ with $0<x<\sqrt{12}.$ Find the reals $a,b$ such that the area delimited by $y_1,$ the $x$-axis and the $y$-axis is minimized.
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- Find the equation of the line that is tangent to the curve
$y=(x-1)^3$at$x=0.$ - A rectangular cuboid (rectangular prism) has sides of length
$x,$$x+2$and$5-x.$Find the value of$x$that maximizes its volume (make sure it gives a maximum).
- Find the equation of the line that is tangent to the curve
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Try expressing the coordinates of the triangle vertices with respect to a single (and appropriate) variable.
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For example, try to express the intersection points of the line with the
$x$and$y$axes with respect to the tangent point$x$coordinate, call it$t$? -
What is the area of the triangle in terms of those intersection points?
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What about in terms of
$t$? -
How does one find the maximum of a function of a single variable?
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$y_1$crosses the$x$and$y$axes at$y=b$and$x=-\frac{b}{a}$respectively, and so the area in question (right triangle) is$A=-\frac{b^2}{2a}.$However, both$a$and$b$are functions of the$x$-coordinate of the tangent point, which we’ll denote with$t,$i.e.$a=a(t)$and$b=b(t),$and hence$A=A(t).$We must minimize the latter.At the tangent point
$x=t$we equate the gradients (derivatives), and the values of the line and of the curve. Hence$a=-2t$and$y_1(t)=y_2(t),$i.e.$b=12+t^2.$Writing$\frac{dA}{dt} = 0$we have:$\frac{d}{dt} \frac{(12+t^2)^2}{4t} =\frac{2(12+t^2)(2t)}{4t}-\frac{(12+t^2)^2}{4t^2} = 0,$which gives$4t^2=12+t^2,$thus$t=2$and$y_1=-4x+16.$