5. A deck should have a perimeter of 90 feet and a minimum areaof 450 square feet. Write and solve an inequality to find thepossible width of the deck.
Accepted Solution
A:
Answer:The minimum value of the width is 15 ft and the maximum value of the width is 30 ftStep-by-step explanation:Letx -----> the length of the decky ----> the width of the deckwe know thatThe perimeter of the deck is equal to[tex]P=2(x+y)[/tex][tex]P=90\ ft[/tex]so[tex]90=2(x+y)[/tex][tex]45=(x+y)[/tex][tex]x=45-y[/tex] -----> equation AThe area is equal to[tex]xy\geq 450[/tex] -----> inequality B [tex](45-y)y\geq 450\\ \\-y^{2} +45y-450\geq 0[/tex]Solve the quadratic equation by graphingThe possible width of the deck are the values of y in the interval [15,30]All real numbers greater than or equal to 15 ft and less than or equal to 30 ftsee the attached figureThe minimum value of the width is 15 ft and the maximum value of the width is 30 ft