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Example: Applications of Radical Functions

A rectangular field has a perimeter of 24 meters with a length of 4 meters and a width of x+2\sqrt{x+2} meters. Find the area of the rectangle.


The perimeter of a rectangle is:
P=2w+2lP=2w+2l
where w = width and l = length ad l,w0l,w\geq0. Thus,
P=2w+2l24=2(x+2)+2(4)16=2x+28=x+282=x+2642=x  x=62\begin{array}{rcl} P&=&2w+2l\\\\ 24&=&2(\sqrt{x+2})+2(4)\\\\ 16&=&2\sqrt{x+2}\\\\ 8&=&\sqrt{x+2}\\\\ 8^2&=&x+2\\\\ 64-2&=&x\\\\ \therefore~~x&=&62 \end{array}

Verify:
24=2(62+2)+2(4)16=262+28=62+28=8\begin{array}{rclc} 24&=&2(\sqrt{62+2})+2(4)\\\\ 16&=&2\sqrt{62+2}\\\\ 8&=&\sqrt{62+2}\\\\ 8&=&8&\checkmark \end{array}

So, the width is x+2=62+2=8\sqrt{x+2}=\sqrt{62+2}=8.

Therefore, the area of the rectangular field is:
A=84=32m2A=8\cdot4=32m^2

Practice: Applications of Radical Functions

Solve for VV
r=3V34πr=\sqrt{\frac{3V^3}{4\pi}}

Practice: Applications of Radical Functions

A garden is in the shape of an isosceles right triangle. The hypotenuse is 10ft. long. The cost of fencing is $1.25/ft. How much will it cost to fence in the garden?
Extra Practice