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​ Quadratics word problem, don’t use functions

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Hello, I need help with this question! I am very stuck but have an idea on what to do! Notice in the question it asks for the “minimum” value of the width, therefore, we must find an equation and find the vertex to determine the minimum width. Please don’t use functions IF YOU can, refer f(x) as y, as we’ve yet to learn about functions. Thank you!

 Dec 8, 2018

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 #1
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\(A=2(15+2w)w+2(25w) = \\ 30w+4w^2+50w = 4(w^2+20w)\\ A \geq 15\cdot 25 = 375\\ 4(w^2+20w) \geq 375\\ 4w^2+80w - 375 \geq 0\\ (2w+20)^2-775\geq 0\\ (2w+20)^2 \geq 775 2w+20 \geq \sqrt{775} \\ w \geq \dfrac{\sqrt{775}-20}{2} = \dfrac{5}{2} \left(\sqrt{31}-4\right)\)

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 Dec 8, 2018
 #2
avatar+115 
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okay I got the same answer in a rather simpler format.

2x-15(2x-25)=750 (750 represents the area of pool + area of sidewalk together)

4x^2 - 80x - 375 = 0 

Quadratic formula magic

and p**f same answer 3.92

Drazil  Dec 9, 2018

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