Question 97846
THIS IS WHY A STRONGLY RECOMMEND YOU DRAW A DIAGRAM BEFORE YOU START TO SOLVE THESE TYPE OF PROBLEMS. YOU SHOULD HAVE DRAWN A 20 BY 30 RECTANGLE.
THIS GIVES YOU A 20*30=600 SQUARE FT.
NOW DRAW A BORDER INSIDE THIS RECTANGLE WITH A STANDARD WIDTH & LABEL THE WIDTH OF THE PATH=X.
NOW SHOULD SEE THAT THE INSIDE AREA THAT=400 SQUARE FT. HAS THE LENGTH OF (30-2X) & HAS A WIDTH OF (20-2X).
NOW SET THE PRODUCT OF THE LENGTH TIMES THE WIDTH EQUAL TO 400.
(30-2X)(20-2X)=400 NOW MULTIPLY THE SIDE DIMENTIONS
600-40X-60X+4X^2=400
4X^2-100X+600-400
4X^2-100+200=0 NOW WXTRACT THE 4 FROM ALL TERMS
4(X^2-25X+50)=0 NOW FACTOR THE EQUATION
USING THE QUADRATIC EQUATION
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
X=(25+-SQRT[-25^2-4*1*50])/281
X=(25+-SQRT[625-200])/2
X=(25+-SQRT425)/2
X=(25+-20.62)/2
X=45.62/2
X=22.81 FT. THIS CANNOT BE THE ANSWER BECAUSE IT IS WIDER THAN THE ENTIRE GARDEN.
X=(25-20.62)/2
X=4.38/2
X=2.19 FT. ANSWER FOR THE WIDTH OF THE PATH.
PLUG IN THESE FIGURES INTO YOUR DIAGRAM FOR THE PROOF.